Stats Test 3 Prep

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 2.75 days with an approximately normal distribution.

(a) =NORM.DIST(6,7.37,2.75,1)-NORM.DIST(5,7.37,2.75,1) (b) =1-NORM.DIST(6,7.37,2.75,1)

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 62% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is

.0125

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?

.0132 inches above the mean

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.

.3694

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 following histogram displays the distribution of battery life (in hours) for a certain battery model used in cell phones: Suppose that battery life is a normal random variable with μ = 8 and σ = 1.2. Using the Standard Deviation Rule, how likely is it that a randomly chosen battery lasts longer than 10.4 hours? 0.50 0.16 0.05 0.025

0.025 Correct. 10.4 is 2 standard deviations above the mean of 8. To calculate probability above this point: 1 − 0.95 = 0.05/2 = 0.025.

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 Correct. Recall that a z-score is calculated as (value − mean)/(standard deviation). So, (31.60 − 28.08)/1.97 = 1.79 (rounded)

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. What is the completion time for the 200-meter backstroke for a female with a z-score of −1.64? (Round answer to 1 decimal place.) A. 130.8 B. 141.0 C. 129.5 D. 152.5

129.5 Correct. A z-score of −1.64 means that the actual time is 1.64 standard deviations below the mean, and therefore the time we are looking for is mean −1.64 SD = 141 − 1.64*7 = 129.52 or 129.5.

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.

159.9 minutes Correct. 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)

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

The number of days (X) after mailout it takes a utility company to receive payment for a customer's bill is a normal random variable with a mean of 31 days and a standard deviation of 6 days. About 16% of bills will be received earlier than -- days after mailout.

25 Correct. This value is 1 standard deviations below the mean of 31 days, so the percent of bills paid earlier than 25 days is 16%. Recall that for a normal random variable 68% of observations will lie within 1 standard deviations of the mean, and 32% will lie outside this range. This amount is split equally between the upper and lower tails of the distribution.

In June 2017, a survey was conducted in which a random sample of 1426 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—27%, No—67%, Unsure—6% Which of the following is true about this scenario?

27%, 67%, and 6% are all statistics.

Pictured below (in scrambled order) are three histograms. One of them represents a population distribution. The other two are sampling distributions of x-bar: one for sample size n = 5 and one for sample size n = 30. Based on the histograms, what is the most likely value of the population mean?

3.0 Correct. We know that the sampling distribution of x-bar has mean μ, the same as the population mean. Since we see from histogram 3 that the sampling distribution of x-bar has a mean of approximately 3, this must also be the mean in the population.

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?

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 _____%. (round to the nearest whole number)

55%

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?

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?

89 We need to find the exam score such that the probability of getting a score above it is 0.04. Equivalently (and more practical, given the way our table works) we need to find the exam score such that the probability of getting a score below it is 1 - 0.04 = 0.96. Looking in the body of the table for the table entry that is closest to 0.96 (which is 0.9599) we learn that the exam score that we are looking for has a z-score of 1.75. This means that the exam score that we are looking for is 1.75 SD above the mean, and therefore is: 75 + 1.75 SD = 75 + 14 = 89.

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 2.75 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.

Correct! The probability that their hospital stay is from 5 to 6 days is: .11478. We can find this by calculating: 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) Part 2 Correct! The probability that their hospital stay is greater than 6 days is: .69082. 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).

A political polling agency wants to take a random sample of registered voters and ask whether or not they will vote for a certain candidate. One plan is to select 400 voters, another plan is to select 1,600 voters. If the study were conducted repeatedly (selecting different samples of people each time), which one of the following would be true regarding the resulting sample proportions of "yes" responses?

Different sample proportions could result each time, but for either sample size, they would be centered (have their mean) at the true population proportion.

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 25 points on this test, rounded to five decimal places? (b) What is the 63 percentile of the distribution of test scores, rounded to three decimal places?

Feedback: Part 1 Incorrect. Don't forget that at least 25 means 25 or more. The correct proportion of students is: .06681. We wish to calculate P(X ≥ 25). 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 1. Part 2 Incorrect. The 63 percentile of the distribution of test scores is: 22.664. We can find the 63th percentile by using the Excel function =NORM.INV(0.63, mean, standard deviation).

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:

HERE THE DISTRIBUTION OF HEIGHT FOLLOWS NORMAL WITH MEAN 175 ANS STANDARD DEVIATION 6 SO P(X<163) WHEN X~N(175,6) P(X<163) IS GIVEN BY THE COMMAND pnorm(163,175,6) WHICH IS 0.02275013 SO P(X<163) IS 0.02275013 SO OUT OF 1000, 1000*0.02275013=22.75013=23 OPTION B 23 ANS.

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 and II only

The average 30- to 39-year old man is 69.5 inches tall, with a standard deviation of 2.7 inches, while the average 30- to 39-year old woman is 64.2 inches tall, with a standard deviation of 3.2 inches. Who is relatively taller based on their comparison to their gender, LeBron James at 81 inches or Candace Parker at 76 inches?

LeBron is relatively taller because he has a larger z-score Correct! Use the z-score formula: (x-mu)/sigma or the Excel function STANDARDIZE(x,mu,sigma) to find the z-scores of Candace and LeBron.

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.

Incorrect. The correct answer is: .9863. We need to first find the standard deviation of the sampling distribution of the sample means by calculating: standard deviation/ sqrt of n then 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 65% 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?

Incorrect. The correct probability is: .00083. We need to first determine the standard deviation of the sampling distribution. To do this, we calculate: √p(1−p)n , where p = .65. Then, we must determine our z-score for determining our normal probability. We calculate: Then, we calculate P(Z>z) by calculating =1- Norm.S.Dist(Z,1).

Left-handedness occurs in about 12% of all Americans. Males are slightly more likely than females to be left-handed, with 13% of males and 11% of females being left-handed. Suppose a random sample of 80 females and 100 males is chosen. Let X be the number of males (out of the 100) who are left-handed. Let Y be the number of females (out of the 80) who are left-handed. Let Z be the total number of left-handed individuals in the sample (males and females together). Which of the following is true regarding the random variables X and Y?

Only X can be well approximated by a normal random variable. Correct. A binomial random variable with parameters n and p can be well approximated by a normal random variable if both n * p ≥ 10 and n * (1 − p) ≥ 10. For X, n = 100, p = 0.13, and 100 * 0.13 and 100*0.87 > 10 → so X can be well approximated by a normal random variable. However, for Y, n = 80, and p = 0.11, and 80 * 0.11 = 8.8 < 10 → cannot be well approximated by a normal random variable.

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.

Part 1 Incorrect. To find the 99th percentile, we can use the Excel function =NORM.INV(0.99,mean, standard deviation). Part 2 Incorrect. Recall, 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). What would that probability be?

Suppose we take repeated random samples of size 20 from a population with a mean of 60 and a standard deviation of 8. Which of the following statements is true about the sampling distribution of the sample mean (x̄)? Check all that apply.

The distribution will be normal as long as the population distribution is normal. The distribution's mean is the same as the population mean 60. Correct!

Concert marketing: GSU's Rialto Center for the Performing Arts wanted to investigate why ticket sales for the upcoming season significantly decreased from last year's sales. The marketing staff collected data from a survey of community residents. Out of the 110 people surveyed, only 7 received the concert brochure in the mail. Which of the following is a reason that the marketing staff should not calculate a confidence interval for the proportion of all community residents who received the concert brochure by mail?

The sample needs to be random, but we don't know if it is. The actual count of community residents who received the concert brochure by mail is too small. n^p is not greater than 10. Correct. The sample needs to be random in order to estimate the population proportion from a sample proportion. Also, since there were fewer than 10 community residents who received the concert brochure by mail, the normal distribution model does not apply. Lastly, in order to use the normal model for the distribution of sample proportions which allows us to calculate a confidence interval, we must have at least 10 actual successes, n^p. In this case, there were only 7 community residents who received the concert brochure by mail.

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

A political polling agency wants to take a random sample of registered voters and ask whether or not they will vote for a certain candidate. One plan is to select 400 voters, another plan is to select 1,600 voters. Which one of the following is true regarding the standard deviation of the sampling distribution of the sample proportion, ^p, of "yes" responses?

The standard deviation of the sampling distribution will be 2 times larger with sample size 400

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. Correct. The question is asking for 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.

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion, ^p, to be within 0.04 of the true proportion, p (i.e., between 0.16 and 0.24)?

There is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

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 --minutes _________--(higherlower) than his --.

This week his time was 2 *standard deviations *lower than his *average time.

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 15 from a population that is normally distributed. We take repeated random samples of size 50 from a population of unknown shape. Correct. The distribution of x-bar will be normal when sampling from a normal population regardless of sample size or when sample size is larger than 30 if sampling from an unknown population.

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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?

sd (standard deviation first)=sqrt(.78*(1-.78)/100) then =1-norm.dist(.80,.78, sd, 1) = .31462 (rounded)

he 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

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