Stats Unit 4 &5

The lifetime in miles for a certain brand of tire is normally distributed with a mean of 22,000 miles and a standard deviation of 3,100 miles The tire manufacturer wants to offer a money-back guarantee so that no more than 3% of tires will qualify for a refund. What is the minimum number of miles the manufacturer should guarantee that the tires will last? 15,800 miles 16,170 miles 25,007 miles 27828

B 16170

In which of the following scenarios would the distribution of the sample mean x-bar be normally distributed? Check all that apply.We take repeated random samples of size 10 from a population of unknown shape. We take repeated random samples of size 15 from a population that is normally distributed. We take repeated random samples of size 50 from a population of unknown shape. We take repeated random samples of size 25 from a population that of unknown shape.

We take repeated random samples of size 50 from a population of unknown shape. We take repeated random samples of size 15 from a population that is normally distributed.

Find probability of the z score

We'll use an Excel function called =NORM.S.DIST(), which takes a z-score and uses it to calculate a probability.

Find mean and stdev N=750 P=.7

X= 525, σX= 12.55 since X is a binomial random variable, μX= np = 750(.7) = 525. σ2X= np(1 - p) = 750(.7)(.3) = 157.5, and therefore σX= sqrt(157.5) = 12.55. (Note that "sqrt" stands for square root.)

The confidence interval is given by

sample mean ± margin of error

Of the 1,019 randomly chosen adults, 60% had used the Internet within the past month to obtain medical information. Use the results of this survey to create an approximate 95% confidence interval estimate for the percentage of all American adults who have used the Internet to obtain medical information in the past month

special p= .6 = SQRT (.6(1-.6)/1019) = .015

T model

standard dev for population is unknown (mean- T * (s/sqrt(N) + t*(s/sqrt(N)

The run times of a marathon runner are approximately normally distributed. The z-score for his run this week is - 2 . Which one of the following statements is a correct interpretation of his z-score? This week his time was 2______ , _________ (higher or lower) than his --(average time, best time, ever time) last week.

stdeviations lower than his average time

The number of hours a light bulb burns before failing varies from bulb to bulb. The distribution of burnout times is strongly skewed to the right. The Central Limit Theorem says that 1.as we look at more and more bulbs, their average burnout time gets ever closer to the mean for all bulbs of this type. 2. the average burnout time of a large number of bulbs has a distribution of the same shape (strongly skewed) as the distribution for individual bulbs. 3. the average burnout time of a large number of bulbs has a distribution that is close to Normal. 4. he average burnout time of any number of bulbs has a distribution of the same shape (strongly skewed) as the distribution for individual bulbs. 5. the average burnout time of any number of bulbs has a distribution that is close to Normal.

the average burnout time of a large number of bulbs has a distribution that is close to Normal.

probability distribution of a random variable can be supplemented with

umerical measures of the center and spread of the random variable.

The grades on a statistics test are normally distributed with a mean of 62 and Q1=52. If the instructor wishes to assign B's or higher to the top 30% of the students in the class, what grade is required to get a B or higher?

you need to determine the standard deviation of the distribution; you can do this by finding the z-score for Q1 and plugging it into σ = (x-μ)/z. By using =Norm.S.Inv(0.25), the z-score of Q1 is -0.6745. Then, you have to solve σ = (x-μ)/z which would be σ = (52-62)/-0.6745 = 14.8260. Finally, you can use the function =Norm.Inv(0.70,62,14.8260) to get value of X such that 70% of the grades are below this value (which corresponds to 30% above). The final answer is 69.77

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points. Suppose that the top 4% of the exams will be given an A+. In order to be given an A+, an exam must earn at least what score? Report your answer in whole numbers.

you need to find the exam score such that the probability of getting a score above it is 0.04. Equivalently, you need to find the exam score such that the probability of getting a score below it is 1 − 0.04 = 0.96. Determine the z-score that has a probability closest to 0.96. The exam score that you are looking for is: μ + z(σ). Answer not 79.1276 and not 83

z-score

z-score = (value - mean)/standard deviation

spread of quantitative data

μ = population mean, σ = population standard deviation ¯x = sample mean μ = σ√n

Using the information from the previous two questions, what is the expected gain or loss for delivering a package?

μX = (5.20)(0.96) + (-14.80)(0.04) = 4.40 This means that in the long run, each package delivered has an expected gain of $4.40.

Spread Categorical

√p(1−p)n stat= p^ Parameter= population proportion

In order to raise money, a charity decides to raffle off some prizes. The charity sells 2,000 raffle tickets for $5 each. The prizes are: 10 movie packages (two tickets plus popcorn) worth $25 each 5 dinners for two worth $50 each 2 smart phones worth $200 each 1 flat-screen TV worth $1,500 What is the expected gain or loss if you buy a single raffle ticket? The expected value can be written as E(X).

$20 net gain if you win the movie package $45 net gain if you win the dinner for two $195 net gain if you win the smart phone $1,495 net gain if you win the TV $5 net loss if you do not win a prize movie package 25 - 5 10 / 2000 dinner for two 50 - 5 5 / 2000 smart phone 200 - 5 2 / 2000 TV 1500 - 5 1 / 2000 nothing 0 - 5 (2000 - 10 - 5 - 2 - 1) / 2000

n September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Congress a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Congress a great deal or a moderate amount. Calculate the 99% confidence interval for the proportion of all American adults who blamed Congress a great deal or a moderate amount for U.S. economic problems.

(0.489, 0.571)

A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a normal distribution with mean 16.05 ounces with a standard deviation of .005 ounces. If four bottles are randomly selected each hour and the number of ounces in each bottle is measured, then 95% of the means calculated should occur in what interval? Hint: the standard deviation rule says that 95% of the observations are within how many standard deviations away from the mean? Round answers to four decimal places.

(16.045,16.055)

The heights of students at a college are normally distributed with a mean of 175 cm and a standard deviation of 6 cm. One might expect in a sample of 1000 students that the number of students with heights less than 163 cm is:

23

Z score

(value-mean)/standard deviation To calculate the z-score, we'll use the equation z = (x − μ)/σ. In this case, we get z=(700−507)/111, or 1.738739.

top 4%

1.5 stdeviations above

Confidence interval from t value

Excel command: =T.INV.2T(0.05,24)

Parameter

A parameter is a number that describes the population; a statistic is a number that is computed from the sample

The insurance company gets information about gas leakage in several houses that use the same gas provider that your customer does. In light of this new information, the probabilities of total loss and 50% damage (that were originally .002 and .008, respectively) are tripled (to .006 for total loss and .024 for 50% damage). Obviously, this change in the probabilities should be reflected in the annual premium, to account for the added risk that the insurance company is taking. What should be the new annual premium (instead of $1,350), if the company wants to keep its expected gain of $750? Guidance: Let the new premium (instead of 1,350) be denoted by N, for new. Set up the new probability distribution of X using the updated probabilities, and using N instead of 1,350. (The answer to question 1 will help.) The question now is: What should the value of N (the new premium) be, if we want the mean of X to remain 750? Set up an equation with N as unknown, and solve for N.

(N - 100,000)(.006) + (N - 50,000)(.024) + N(.97)750 = (N - 100,000)(.006) + (N - 50,000)(.024) + N(.97) Thus, 750 = N - 600 - 1,200, or N - 1,800. And therefore, N = 750 + 1,800 = 2,550. the premium changes from $1,350 to $2,550.

Household size in the United States has a mean of 2.6 people and standard deviation of 1.4 people. What is the probability that the mean size of a random sample of 100 households is more than 3?

. Its mean is the same as the population mean, 2.6, and its standard deviation is the population standard deviation divided by the square root of the sample size: σ√n= 1.4√100=.14. The z-score for 3 is 3−2.61.4√100=0.40.14=2.86 The probability of the mean household size in a sample of 100 being more than 3 is therefore P(¯¯¯X > 3) = P(z > 2.86) = P(z < -2.86) = .0021.

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule. (a) What proportion of the students scored at least 27 points on this test, rounded to five decimal places? (b) What is the 42 percentile of the distribution of test scores, rounded to three decimal places?

.00621 pt 2 21.596. We can find the 42th percentile by using the Excel function =NORM.INV(0.42, mean, standard deviation).

A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let ^p denote the proportion in the sample who say they support the increase. Suppose that 75% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is

.012

A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let ^p denote the proportion in the sample who say they support the increase. Suppose that 54% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is

.0129

A factory produces plate glass with a mean thickness of 4mm and a standard deviation of 1.1mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 3.83 mm?

.06112. 1. find standard deviation by deviation / SQRT(n) 2. Calculate z score

The distribution of the diameters of a particular variety of oranges is approximately normal with a standard deviation of 0.3 inch. How does the diameter of an orange at the 67th percentile compare with the mean diameter? 0.201 inches below the mean 0.132 inches below the mean 0.440 inches above the mean 0.132 inches above the mea.132 inches above the mean

.132 above

The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3. Suppose a golfer played the course today. Find the probability that her score is at least 74 0.4772 0.1293 0.0228 0.3694

.3694

Suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. Using the Standard Deviation Rule, what is the probability that a randomly chosen battery will last between 6.8 and 9.2 hours? 0.50 0.68 0.95 0.997

.68 determine how many standard deviations away from the mean the value 6.8 and 9.2 are and then use the standard deviation rule to determine the probability

A factory produces plate glass with a mean thickness of 4mm and a standard deviation of 1.1mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 4.06 mm?

.70728 =stdev/SQRT(n) z= x - mean / stdev(SQRT(N))

A factory produces plate glass with a mean thickness of 4mm and a standard deviation of 1.1mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 4.13 mm?

.88136

The lifetime in miles for a certain brand of tire is normally distributed with a mean of 22,000 miles and a standard deviation of 3,100 miles The tire manufacturer wants to offer a money-back guarantee so that no more than 3% of tires will qualify for a refund. What is the minimum number of miles the manufacturer should guarantee that the tires will last?

16170

Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120. Suppose you take a simple random sample (SRS) of 25 students from this distribution. What is the probability that a SRS of 25 students will spend an average of between 600 and 700 dollars? Round to five decimal places

.96278. We can standardize both of these values by calculating their z-scores. Then, we use the following in Excel = Norm.S.Dist(Z-score for 700,1) - Norm.S.Dist(Z-score for 600, 1). Alternatively, we can calculate, =Norm.Dist(700, Pop. mean, standard deviation of sample distribution, 1) - Norm.Dist(600, Pop. mean, standard deviation of sample distribution, 1)

Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120. Suppose you take a simple random sample (SRS) of 35 students from this distribution. What is the probability that a SRS of 35 students will spend an average of between 600 and 700 dollars? Round to five decimal places

.9863. = stdev / sqrt(n) P(600< x) . We can standardize both of these values by calculating their z-scores. Then, we use the following in Excel = Norm.S.Dist(Z-score for 700,1) - Norm.S.Dist(Z-score for 600, 1). Alternatively, we can calculate, =Norm.Dist(700, Pop. mean, standard deviation of sample distribution, 1) - Norm.Dist(600, Pop. mean, standard deviation of sample distribution, 1)

A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let ^p denote the proportion in the sample who say they support the increase. Suppose that 17% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is

0.0097.

Which of the following statements about the sampling distribution of the sample mean, x-bar, is true? Check all that apply.

1 . The distribution is normal regardless of the sample size, as long as the population distribution is normal. 2. The distribution's mean is the same as the population mean. 3. The distribution's standard deviation is smaller than the population standard deviation. 4. The distribution is normal regardless of the shape of the population distribution, as long as the sample size, n, is large enough.

According to the Bureau of Labor Statistics, in May 2014 the mean hourly wage of a social worker was 28.08 with a standard deviation of 1.97. Jeronica is a social worker with an hourly wage of 31.60. What is Jeronica's z-score? (Round your answer to 2 decimal places.)

1.79

Scores on the math portion of the SAT (SAT-M) in a recent year have followed a normal distribution with mean μ = 507 and standard deviation σ = 111. What is the probability that the mean SAT-M score of a random sample of 4 students who took the test that year is more than 600? Explain why you can solve this problem, even though the sample size (n = 4) is very low.

111/SQRT(4) = 55.5 z score of 600= 600-507 / 111/SQRT(4)= 1.68 P(Z>1/68)= symmetry= P(Z<-1.68)= .0465

A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a normal distribution with mean 16.05 ounces with a standard deviation of .3005 ounces. If four bottles are randomly selected each hour and the number of ounces in each bottle is measured, then 95% of the means calculated should occur in what interval? Hint: the standard deviation rule says that 95% of the observations are within how many standard deviations away from the mean? Round answers to four decimal places.

15.7495, 16.3505.

A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a normal distribution with mean 16.05 ounces with a standard deviation of .3 ounces. If four bottles are randomly selected each hour and the number of ounces in each bottle is measured, then 95% of the means calculated should occur in what interval? Hint: the standard deviation rule says that 95% of the observations are within how many standard deviations away from the mean? Round answers to four decimal places.

15.75,16.35

The time it takes a surgeon to complete a laproscopic surgery to remove the gall bladder is normally distributed with a mean of 132.4 minutes and a standard deviation of 15.7 minutes. A patient's risk of complications is increased the longer he is in surgery. An unusually risky surgery is one that is in the top 4% of all surgery lengths. What is the minimum surgery length (in minutes) that would be considered unusually risky? Note: Answers are rounded to 1 decimal place.247.4 minutes 104.9 minutes 147.5 minutes 159.9 minutes

159.9 You recognized that the boundary for the top 4% is the same as the boundary for the bottom 96% and you used the Excel formula: =NORM.INV(0.96,132.4,15.7)

You conduct a random experiment in which you toss a coin 10 times. How many possible outcomes with exactly 6 heads are there in this random experiment?

210 Indeed, 10! / 6!(10-6)! = (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10) / (1 * 2 * 3 * 4 * 5 * 6)(1 * 2 * 3 * 4) = 210.

In June 2017, a survey was conducted in which a random sample of 1490 U.S. adults was asked the following question: "In 1973 the Roe versus Wade decision established a woman's constitutional right to an abortion, at least in the first three months of pregnancy. Would you like to see the Supreme Court completely overturn its Roe versus Wade decision, or not?" The results were: Yes—28%, No—61%, Unsure—11% Which of the following is true about this scenario? 28%, 61%, and 11% are all statistics. If another random sample of size 1490 U.S. adults were to be chosen, we would expect to get the exact same distribution of answers. 28%, 61%, and 11% are all parameters.

28%, 61%, and 11% are all statistics.

X is binomial with n = 300 and p = .9, and would therefore be approximated by a normal random variable having mean μ = 300 * 0.9 = 270 and standard deviation σ = sqrt(300 * 0.9 * 0.1) = sqrt(27) = 5.2. Note that if you look at the histogram, this makes sense. The distribution is indeed centered at 270, and extends approximately ___ deviations

3 standard deviations (3 * 5.2 = 15.6) on each side of the mean

In a study of the effects of acid rain, a random sample of 100 trees from a particular forest is examined. Forty percent of these show some signs of damage. Which of the following statements is correct? Less than 40% of the trees in the forest show some signs of damage More than 40% of the trees in the forest show some signs of damage 40% of all trees in the forest show some signs of damage 40% is a statistic 40% is a parameter

40% is a statistic

The weights of cockroaches living in a typical college dormitory are approximately normally distributed with a mean of 80 grams and a standard deviation of 4 grams. The percentage of cockroaches weighing between 77 grams and 83 grams is about

55 percent

Probability Notation

68% that X falls within 1 σ of μ , that is, in the interval μ±σ 95% that X falls within 2 σ of μ , that is, in the interval μ±2σ 99.7% that X falls within 3 σ of μ , that is, in the interval μ±3σ 0.68=P(μ−σ<X<μ+σ) 0.95=P(μ−2σ<X<μ+2σ) 0.997=P(μ−3σ<X<μ+3σ)

Z value for confidence intervals

90% confidence interval: zc≈ 1.645 95% confidence interval: zc≈ 1.960 (2 is a rough approximation; 1.960 is more accurate) 99% confidence interval: zc≈ 2.576

standard deviation

=SQRT(variance)

Variance

=SUMPRODUCT(1st arrayx)^2*Y

Using Excel to find tc for different values of n where the degrees of freedom df=n−1:

=T.INV.2T(probability, deg_freedom), for the 95% C.I. discussed above:

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rulea) What proportion of the students scored at least 29 points on this test, rounded to five decimal places?. (b) What is the 20 percentile of the distribution of test scores, rounded to three decimal places?

A. .00023 b.The 20 percentile of the distribution of test scores is: 20.317. We can find the 20th percentile by using the Excel function =NORM.INV(0.20, mean, standard deviation).

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule (a) What proportion of the students scored at least 21 points on this test, rounded to five decimal places? (b) What is the 23 percentile of the distribution of test scores, rounded to three decimal places?

A. .69146. We wish to calculate P(X ≥ 21). Since the normal distribution is continuous, we do not have to worry about the inequality signs. Excel will only calculate < or ≤ so we need to re-write the problem by using the complement rule. Thus, we can can use the Excel function = 1-Norm.Dist(x, mean, standard deviation, 1). The normal distribution will ALWAYS have a cumulative equal to 1B. 2. The 23 percentile of the distribution of test scores is: 20.522. We can find the 23th percentile by using the Excel function =NORM.INV(0.23, mean, standard deviation)

Z score excel

Alternatively, we can use the =STANDARDIZE function in Excel. To determine the standardized value, enter =STANDARDIZE(x,μ,σ) =STANDARDIZE(700,507,111)

. According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was $2,600. Assume that the standard deviation in Pell grants awards was $500. If we randomly sample 36 Pell grant recipients, would you be surprised if the mean grant amount for the sample was $2,940? Pick the correct response that gives the best reason. 1. No, 2,940 would not be surprising because this sample result is only 340 greater than the overall mean grant amount of 2,600, and we expect there to be variability in sample means 2. .Yes, 2,940 would be surprising because this sample result is more than 3 standard deviations from the overall mean grant amount of 2,600. 3. Yes, 2,940 would be surprising because this sample result is 340 greater than the overall mean grant amount of 2,600.No, 2,940 would not be surprising because th 4. is sample result is within 2 standard deviations of the overall mean grant amount of 2,600.

Answer; 2 For n=36, sample means are approximately normal, so we can use the Standard Deviation Rule. Three standard deviations above 2,600 is 2,600 + 3(500/6)) = 2,850). So 2,940 is more than 3 standard deviations above 2,600, thus this sample mean would be surprising

A distribution of a single statistic from repeated random samples of the same size from the same population refers to which of the following? Distribution of summary statistics The normal curve Sampling distribution of a statistic Distribution of population parameters Random sampling

Sampling distribution of a statistic

A multiple-choice quiz has 15 problems, each with 5 possible answers, only one of which is correct. A student who does not attend lectures has no clue, and uses an independent random guess to answer each of the problems. The random variable X is the number of questions the student got right.

Binomial This experiment consists of 15 trials (each quiz question is a trial). Since the student uses an independent random guess to answer each of the 15 questions, the trials are independent, and all have the same probability of being correctly answered. X represents how many of the 15 questions ended in a "success" (were answered correctly)

In the following random experiment, decide whether the random variable X is binomial or not. It is known that roughly 8% of males have some sort of color vision deficiency (also known as color-blindness). A random sample of 1,000 males was selected. Let X be the number of males out of the sample that are color-blind.

Binomial This random experiment consists of 1,000 trials (1,000 males), all having the same probability of being color-blind ("success"), and X represents how many of the trials (subjects) ended in a "success" (are color-blind). Note that even though the selection here is without replacement, we can disregard the dependence problem, and regard the trials as independent, since the population (all males) is more than 10 times the sample size (10 * 1,000 = 10,000).

According to the National Post secondary Student Aid Study conducted by the U.S. Department of Education in 2008, 62% of graduates from public universities had student loans. We randomly sample college graduates from public universities and determine the proportion in the sample with student loans. For which of the following sample sizes is a normal model a good fit for the sampling distribution of sample proportions? 102030Both 20 and 30None of these

Correct. Both conditions are met. np = (30)(0.62) = 18.6 and n(1 - p) = (30)(0.38) = 11.4. Both are greater than 10. So a normal model is a good fit for the sampling distribution of sample proportions

Imagine that you have a very large barrel that contains tens of thousands of M&M's. According to the official M&M website, 20% of the M&M's produced by the Mars Corporation are orange. 5 students each take a random sample of 50 M&M's and record the percentage of orange in each sample. Which sequence is the most plausible for the percentage of orange candies obtained in these 5 samples? 20%, 20%, 20%, 20%, 20% 15%, 25%, 22%, 20%, 28% 5%, 80%, 8%, 65%, 70% Any of the above

Correct. We expect p-hats within 1 standard deviation of p = 0.20 to be most common. The standard deviation is about 0.06. 4 of the 5 p-hats in this sequence are within 0.06 of p = 0.20.

Suppose two students from Georgia State University, working as interns for the American National Elections Studies agency (ANES), are both assigned to survey a random sample of registered voters and ask whether or not they will vote for a certain candidate. The first intern plans to select 500 voters and the second intern plans to select 1500 voters. If each intern conducted the study repeatedly (selecting different samples of people each time ... but using the same sample size), which one of the following would be true regarding the resulting sample proportion, ^p, of "yes" responses for each intern? A. For either sample size, using the same size each time, as long as the sampling is done with replacement, their mean would be 0. B. Different sample proportions, ^p, could result for each intern, but for the intern using a sample size of 1500 they would be centered (have their mean) at the true population proportion,p, whereas for sample size 500 they would not. C. Different sample proportions, ^p, could result for each intern, but for sample size 500 they would be centered (have their mean) at the true population proportion, p, whereas for sample size 1500 they would not. D. Different sample proportions, ^p, could result for each intern, but for either sample size, they would be centered (have their mean) at the true population proportion, p.

Different sample proportions, ^p, could result for each intern, but for either sample size, they would be centered (have their mean) at the true population proportion, p.ue to sampling variability, for different samples we will get different values of ^p. S ince the mean of the sampling distribution of ^p is p, the population proportion, whether we take a sample of size 100 or a sample of size 900, the resulting values of ^p will be centered around p in both cases.

special case of the Central Limit Theorem.

For categorical variables, our claim that sample proportions are approximately normal for large enough n i

The following displays two normal distributions. Which of the following are true? I. The mean of A is less than the mean of B. II. The standard deviation of A is less than B. III. The area under the curve of A is less than B.

I & II

There are 6 members in a family (2 parents and 4 children). Let X be the number of family members who have blue eyes. binomial or not binomial

Not binomial The random experiment consists of 6 trials (6 family members), where the outcome of interest (success) is having blue eyes. Even though X represents the number of "successes" among the 6 "trials," X is not binomial, because eye color is hereditary, and therefore the 6 trials are not independent. Knowing, for example, that the father has blue eyes has an impact on the probability that the children have blue eyes. X, therefore, is not binomial.

The estimated standard error is 0.0158 so the confidence interval is given by: 0.53 ± 2.576 (0.0158).What does the 99% confidence level in the previous problem tell us?

Of confidence intervals with this margin of error, 99% will contain the population proportion.

According to the College Board website, the scores on the math part of the SAT (SAT-M) in a certain year had a mean of 507 and standard deviation of 111. Assume that SAT scores follow a normal distribution. One of the criteria for admission to a certain engineering school is an SAT-M score in the top 2% of scores. How does this translate to an actual SAT-M score? In other words, how high must a student score on the SAT-M in order for his application to be considered? A different way to ask the same question is "What is the 98th percentile of the SAT-M distribution?"What value of x do we need to use with the normal table in order to solve this problem?

P(X <x) = 0.98

With previous case, find P(X ≥ 31). For the approximation to be better, use the continuity correction as we did in the last example. In other words, rather than approximating P(X ≥ 31) by P(Y ≥ 31), approximate it by P(Y ≥ 30.5).

P(X ≥ 31) ≈ (normal approximation + continuity correction) ≈ P(Y ≥ 30.5) = P(Z ≥ (30.5-22.5)/4.5) = P(Z ≥ 1.78) = (symmetry) = P(Z ≤ -1.78) = (table) = 0.0375.

integrals

P(a≤X≤b)=(area between a and b and below the density curve)=∫baf(x)dx , where f(x) represents the density curve.

Excel Distribution example

P(−1 < Z < +1) = P(Z < +1) − P(Z < −1) = 0.8413 - 0.1587 = 0.6826. USING EXCEL: to find P(−1<Z<+1) type: =NORM.S.DIST(1,1)-NORM.S.DIST(-1,1)

The Acme Shipping Company has learned from experience that it costs $14.80 to deliver a small package overnight. The company charges $20 for such a shipment, but guarantees that they will refund the $20 charge if it does not arrive within 24 hours. Suppose Acme successfully delivers 96% of its packages within 24 hours. What are the probabilities that correspond to the values for X you found in the previous question?

Since they deliver within 24 hours with a probability of 0.96, 0.96 corresponds to 5.20. Not delivering within 24 hours is the complement of delivering within 24 hours. Thus, the probability corresponding to -14.80 is 1 - 0.96 = 0.04.

The critical T-value is 1.667. The sample size is 71, so we will set df = 70. We want a 90% confidence interval, so we will set the central area to 0.90 Find the margin of error for this 90% confidence interval.

So the margin of error is given by 1.667(3.78 / √71) = 0.75.

Suppose that 78% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patients, what is the probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places?

The correct probability is: .31462. We need to first determine the standard deviation of the sampling distribution. To do this, we calculate: √p(1−p)n , where p = .78. Then, we must determine our z-score for determining our normal probability.

Suppose that 88% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patients, what is the probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places?

The correct probability is: .99309. We need to first determine the standard deviation of the sampling distribution. To do this, we calculate: √p(1−p)n , where p = .88. Then, we must determine our z-score for determining our normal probability. We calculate

X is binomial with n = 225 and p = 0.1. Explain why we can use the normal approximation in this case, and state which normal distribution you would use for the approximation

The normal approximation is appropriate, since the rule of thumb is satisfied: np = 225 * 0.1 = 22.5 > 10, and n(1 − p) = 225 * 0.9 = 202.5 > 10. binomial random variable X by the random variable Y having a normal distribution with mean μ = 225 * 0.1 = 22.5 and standard deviation σ = √(225 * 0.1 * 0.9) = sqrt(20.25) = 4.5

A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 1.25 days with an approximately normal distribution. (a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places.

The probability that their hospital stay is from 5 to 6 days is: .10756. We can find this by calculating: a. P(5<x<6)=P(x<6)-P(x<5) In Excel, we do this by using the function =Norm.Dist(6,mean, standard deviation, 1) - Norm.Dist(5,mean, standard deviation ,1) b. .86346 greater than 6 days is: .86346. We wish to find P(X>6). Excel can only calculate < or ≤ when determining normal probabilities. Thus, we can solve it by calculating =1 - norm.dist(x,mean, standard deviation ,1).

Suppose the American National Elections Studies agency (ANES) wishes to conduct a survey. It plans to ask a yes/no question to determine if those surveyed plan to vote for a certain candidate. One proposal is to randomly select 400 people and another proposal is to randomly select 1600 people. Which of the following is true regarding the sample proportion ^p of "yes" responses? The sample proportion from the sample of 400 is more likely to be close to the true population proportion, p. The sample proportion from sample of 1,600 is more likely to be close to the true population proportion, p. The sample proportion, ^p, in either proposal is equally likely to be close to the true population proportion, p, since the sampling is random.

The sample proportion from sample of 1,600 is more likely to be close to the true population proportion, p

The annual salary of teachers in a certain state X has a mean of $54,000 and standard deviation of σ = $5,000. What is the probability that the mean annual salary of a random sample of 64 teachers from this state is less than $52,000?

The z-score of 52,000 is (52,000 − 54,000)/5000/√(64),= -3.2

Suppose that a candy company packages a bag of jelly beans whose weight is supposed to be 30 grams, but in fact, the weight varies from bag to bag according to a normal distribution with mean μ = 30 grams and standard deviation σ = 3 grams. If the company sells the jelly beans in packs of 9 bags, what can we say about the likelihood that the average weight of the bags in a randomly selected pack is 2 or more grams lighter than advertised?

There is about a 2.5% chance of this occurring. P(x̄ ≤ 28). The distribution of x-bar is normal with a mean = 30g and standard deviation = 3/√(9) = 1. So P(x̄ ≤ 28) = P(z ≤ 2) = 0.025, so there is about a 2.5% chance that the mean bag weight will be less than 28g

Juries should have the same racial distribution as the surrounding communities. According to the U.S. Census Bureau, 18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American. How likely is the resulting sample proportion to be between 0.066 and 0.294 (i.e., 6.6% to 29.4% African American)? a. There is roughly a 68% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion. b. There is roughly a 95% chance that the resulting sample proportion, ^p, will be between 0.066 and 0.294 of the true proportion, p. c. It is certain that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion. d. There is roughly a 99.7% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion. Feedback

There is roughly a 99.7% chance that the resulting sample proportion will be between 0.066 and 0.294 of the true proportion.

Probability using excel

USING EXCEL (instead of the Standard Normal Table) to find the areas under the curve: For example, to find: P(Z<-2.21) in Excel type: =NORM.S.DIST(-2.21,1) Note:If you compare the Excel result to the table result from above, you will see that Excel offers more decimal place accuracy. P(Z<-2.15) in Excel type: =NORM.S.DIST(-2.15,1) P(Z<-2.63) in Excel type: =NORM.S.DIST(-2.63,1)

Male foot lengths have a normal distribution, with μ=11,σ=1.5 inches. What is the probability of a foot length of more than 13 inches?

USING EXCEL: To find P(X>13) type: =1-NORM.DIST(x,μ, σ, 1), specifically, type: =1-NORM.DIST(13,11, 1.5, 1)

Suppose the time to complete a 200-meter backstroke swim for female competitive swimmers is normally distributed with a mean μ = 141 seconds and a standard deviation σ = 7 seconds. Suppose the fastest 6% of female swimmers in the nation are offered college scholarships. In order to be given a scholarship, a swimmer must complete the 200-meter backstroke in no more than how many seconds? Give your answer in whole numbers.

You need to find the time such that the probability of swimming faster (less than the time) is 0.06. Determine the z-score that has a probability closest to 0.06. The time that you are looking for is then calculated as μ + z(σ) = 141 + z * 7.

A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 0.75 days with an approximately normal distribution. (a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places. (b) Find the probability that their hospital stay is greater than 6 days, rounded to five decimal places.

a. .03309 b. .96613

The Environmental Protection agency requires that the exhaust of each model of motor vehicle be tested for the level of several pollutants. The level of oxides of nitrogen (NOX) in the exhaust of one light truck model was found to vary among individually trucks according to a Normal distribution with mean 1.45 grams per mile driven and standard deviation 0.40 grams per mile .(a) What is the 99th percentile for NOX exhaust, rounded to four decimal places? (b) Find the interquartile range for the distribution of NOX levels in the exhaust of trucks rounded to four decimal places

a. 2.3805 =NORM.INV(0.99, mean, standard deviation). b. .5396 The IQR = Q3 - Q1, where Q3 is the third quartile and Q1 is the first quartile. In Excel, Q1 = NORM.INV(0.25, mean, standard deviation). Q3 = NORM.INV(0.75, mean, standard deviation).

Even if you don't get the right answer, the process of trying is what is valuable. Learning is

about the process of reasoning through confusion to understand something new.

he probability of having more than two children is (X > 2) how to find it?

calculate 1-P(X≤2)

The population mean for Verbal IQ scores is 100, with a standard deviation of 15. Suppose a researcher takes 50 random samples with 30 people in each sampleWhat is the standard deviation of the sample means?

he standard deviation of the sample means is calculated by dividing the population standard deviation by the square root of the sample size; therefore, σ/sqrt(n) = 15/sqrt(30)= 15/5.48= 2.74,

o find probabilities of the type P(X = k) or P(X ≤ k) in Excel, we'll use the following function: where

k is the number of successes in trials n is the number of independent trials p is the probability of success in each trial cumulative is a logical value that determines the form of the function. If cumulative is TRUE, then BINOMDIST returns the cumulative distribution function, which is the probability that there are at most k successes: P(X ≤ k). If FALSE, it returns the probability mass function, which is the probability that there are exactly k successes: P(X = k). As practice, follow these steps to find P(X = 4) for our example (where n = 10 and p = .2), and verify that you get the same answer as you did in the last question, where you did it by hand. Now use Excel to find the probability that the student gets no more than 4 questions right: P(X ≤ 4).

larger samples have

less variability than small samples

X is binomial with n = 100 and p = .75, and would therefore be approximated by a normal random variable having mean and standard deviation of

mean μ = 100 * 0.75 = 75 and standard deviation σ = sqrt(100 * 0.75 * 0.25) = sqrt(18.75) = 4.33.

Center: The center of a random variable is measured by its

meanbe interpreted as its long run average.

When the population is not normally distributed, the sampling distribution of the mean approximates which of the following? A normal distribution given a large enough sample A normal distribution A slight positive skew A distribution that is not normal

normal distribution given a large enough sample

The deciles of any distribution are the points that mark

off the highest and lowest 10% of the observations.

Standard Deviation Table

pg 133

A discrete random variable is summarized by its

probability distribution—a list of its possible values and their corresponding probabilities. The sum of the probabilities of all possible values must be 1. The probability distribution can be represented by a table, histogram, or formula.