Exam 2
Suppose the mean income of firms in the industry for a year is 85 million dollars with a standard deviation of 9 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 99 million dollars? Round your answer to four decimal places.
2nd, VARS, NormalCDF Lower: -1E99 Upper: 99 μ= 85 σ= 9 =.9401
Find the area under the standard normal curve to the right of z=−2.79 Round your answer to four decimal places, if necessary.
2nd, VARS, NormalCDF Lower: -2.79 Upper: 1E99 μ= 0 σ= 1 =.9974
The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1400lbs. Find the probability that the weight of a randomly selected steer is between 1539 and 1679lbs. Round your answer to four decimal places.
2nd, VARS, NormalCDF Lower: 1539 Upper: 1679 μ= 1400 σ= 200 =.162
A soft drink machine outputs a mean of 25 ounces per cup. The machine's output is normally distributed with a standard deviation of 4 ounces. What is the probability of overfilling a 29 ounce cup? Round your answer to four decimal places.
2nd, VARS, NormalCDF Lower: 29 Upper: 1E99 μ= 25 σ= 4 =.1587
Suppose the mean income of firms in the industry for a year is 85 million dollars with a standard deviation of 9 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn more than 99 million dollars? Round your answer to four decimal places.
2nd, VARS, NormalCDF Lower: 99 Upper: 1E99 μ= 85 σ= 9 =.05991
A certain insecticide kills 70% of all insects in laboratory experiments. A sample of 10 insects is exposed to the insecticide in a particular experiment. What is the probability that exactly 3 insects will die? Round your answer to four decimal places.
(binomPDF) 2nd, VARS, binompdf trials= 10 p= 0.7 x value= 3 = .009
A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 40% of this population prefers the color green. If 12 buyers are randomly selected, what is the probability that exactly 2 buyers would prefer green? Round your answer to four decimal places.
(binomPDF) 2nd, VARS, binompdf trials= 12 p= 0.4 x value= 2 = .0639
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X=3) n=16 p=0.5
(this is a binomPDF question bc of the = sign) 2nd, VARS, binompdf trials= 16 p= 0.5 x value= 3 (KEEP THE SAME) =.0085
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X≤1) n=5 p=0.6
(this is binomCDF bc of the ≤ sign) 2nd, VARS, binomcdf trials= 5 p= 0.6 x value= 1 (KEEP THE SAME) =.0870
A real estate agent has 16 properties that she shows. She feels that there is a 50% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling less than 5 properties in one week. Round your answer to four decimal places.
(this is binomCDF) 2nd, VARS, binomcdf trials= 16 p= .5 x value= 4 (SUBTRACT 1) =.0384 DONT SUBTRACT 1 FROM TOTAL
A real estate agent has 18 properties that she shows. She feels that there is a 40% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling more than 3 properties in one week. Round your answer to four decimal places.
(this is binomCDF) 2nd, VARS, binomcdf trials= 18 p= 0.4 x value= 3 (KEEP THE SAME) =.03278 1-.03278 = .96722
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X<5) n=6 p=0.3
(this is binomCDF) 2nd, VARS, binomcdf trials= 6 p= 0.3 x value= 4 KEEP X VALUE THE SAME
A real estate agent has 16 properties that she shows. She feels that there is a 50% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling at least 5 properties in one week. Round your answer to four decimal places.
(this is binomCDF) 2nd, VARS, binomcdf trials= 16 p= 0.5 x value= 4 (SUBTRACT 1) =.0384 1-.0384 = .9616
Suppose 46% of politicians are lawyers. If a random sample of size 607 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by more than 4%? Round your answer to four decimal places.
***** 1-***** 2nd VARS normalCDF lower: .46-.04 upper: .46+.04 μ= .46 σ= √.46(1-.46)/607 = .048
Find the area under the standard normal curve to the left of z=0.33 and to the right of z=2.38. Round your answer to four decimal places, if necessary.
****** 1- 2nd, VARS, NormalCDF ****** Lower: .33 Upper: 2.38 μ= 0 σ= 1 =.63796 (DONT FORGET TO SUBTRACT THE 1 IN FRONT!!)
Find the mean of the sampling distribution of sample means using the given information. Round to one decimal place, if necessary. μ=52 σ=9 n=49
52
how to find confidence interval for a population parameter on a TI-84
STAT, tests, 1-PropZint
A student reads that the sample mean ACT math score from a sample of 37 Kansas high school students in 2004 is 21.4, with a sample standard deviation of 2.7. Which of the following gives a 95% confidence interval for the true population mean ACT math score? (Assume the population std is unknown) A (20.50, 22.30) B (19.64, 23.16) C (19.57, 23.23) D (21.10, 21,70)
STAT, tests, Tinterval x bar: 21.4 Sx: 2.7 n: 37 C-level: .95 =(20.50, 22.30) A
A certain test preparation course is designed to help students improve their scores on the GMAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 3 students' scores on the exam after completing the course: 17, 20, 26 Using these data, construct a 95% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Calculate the sample mean for the given sample data. Round your answer to one decimal place. Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Construct the 95% confidence interval. Round your answer to one decimal place.
Sample mean = STAT, Edit, put #s in, 1 VarStats x bar= 21 stdev = Sx = 4.6 critical value: First do: 1-confidence interval/2 (1-.95/2=.025 Next: do: n-1 (3-1=2) 2nd, VARS, invT area: .025 df: 2 =4.303 95% confidence interval: STAT, tests, Tinterval plug in numbers from above = (9.573, 32.427)
In the previous problem, suppose the random variable X represents the number of people who say YES in the sample of 15. Find the mean and standard deviation of X.
mean = np (15)(.2) = 3 mean = 3 stdev = √np(1-p) = √15(.2)(1-.2) = 1.549 stdev = 1.549
fewer degrees o freedom has ___________ area under the tails of the curve
more
A college basketball player has probability 0.50 of making any given shot. Her successive shots are independent. In overtime of an important game, she misses all three shots that she takes. Compute the probability of her missing all three shots. Explain why this does not mean that she "choked," because it could happen by chance with reasonable probability.
n = 3 p = .50 (.50)^3 = .125
binompdf
single binomial probability
bigger degrees of freedom makes the t-distribution approach the shape of the ______________
standard normal
Suppose that a study of elementary school students reports that the mean age at which children begin reading is 5.3 years with a standard deviation of 1.0 years. Step 1 of 2 : If a sampling distribution is created using samples of the ages at which 34 children begin reading, what would be the mean of the sampling distribution of sample means? Round to two decimal places, if necessary.
.34 .17
What value of z divides the standard normal distribution so that half the area is on one side and half is on the other? Round your answer to two decimal places.
0.00
In the previous problem, again let X represent the number of people who say YES in the sample of 15. Find the probability that 10 or more of those sampled say YES.
1 - binomcdf (15, .2, 9) = 1.132E-4 =.000113
discrete random variable
a random variable that may assume either a finite number of values or an infinite sequence of values (gaps)
interval estimate
a range of possible values for a population parameter
The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.48 millimeters and a standard deviation of 0.07 millimeters. Find the two diameters that separate the top 3% and the bottom 3%. These diameters could serve as limits used to identify which bolts should be rejected. Round your answer to the nearest hundredth, if necessary.
1-.03 = .97 2nd, VARS, 3:invNorm area: .97 μ= 5.48 σ= 0.07 =5.61 AND 2nd, VARS, 3:invNorm area: .03 μ= 5.48 σ= 0.07 =5.35 (5.61, 5.35)
Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.4 A university plans to admit students whose scores are in the top 45%. What is the minimum score required for admission? Round your answer to the nearest tenth, if necessary.
1-.45= .55 2nd, VARS, 3:invNorm area: .55 μ= 21.2 σ= 5.4 =21.9
A researcher examines 27 water samples for mercury concentration. The mean mercury concentration for the sample data is 0.097 cc/cubic meter with a standard deviation of 0.0074. Determine the 90% confidence interval for the population mean mercury concentration. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Construct the 90% confidence interval. Round your answer to three decimal places.
1-.90= .05 27-1= 26 2nd, VARS, InvT area: .05 df: 26 =1.706 (you'll get a negative but ignore the negative sign) STAT, TInterval, x bar: .097 Sx: .0074 n= 27 c-level: .90 =(.09457, .09943)
Find the value of z such that 0.9282 of the area lies between −z and z. Round your answer to two decimal places.
1-0.9282 = .0718 .0718/2 = .0359 2nd, VARS, 3:invNorm area: .0359 μ= 0 σ= 1 =-1.8
How to know if a distribution is a probability distribution, it must have 2 conditions:
1. The sum of all of the probabilities must equal 1 2. The probability of any value must be between 0 and 1, inclusively
The mean GPA for 112 residents of the local apartment complex is 1.7. What is the best point estimate for the mean GPA for all residents of the local apartment complex?
1.7
Given the following confidence interval for a population mean, compute the margin of error, E. 10.63<μ<13.19
1/2(13.19-10.63) = 1.28
point estimate
a single estimate of a population parameter - a point on a number line
A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 20.3 pounds and a standard deviation of 3.7 pounds. If a sampling distribution is created using samples of the amounts of weight lost by 94 people on this diet, what would be the mean of the sampling distribution of sample means? Round to two decimal places, if necessary. If a sampling distribution is created using samples of the amounts of weight lost by 94 people on this diet, what would be the standard deviation of the sampling distribution of sample means? Round to two decimal places, if necessary.
20.3 3.7/√94 = 3.8
Suppose 50% of politicians are lawyers. If a random sample of size 853 is selected, what is the probability that the proportion of politicians who are lawyers will be less than 46%? Round your answer to four decimal places.
2nd VARS normalCDF lower: -1E99 upper: .46 μ= .5 σ= √.5(1-.5)/853 =.0097
Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation of 0.5 feet. A sample of 51 men's step lengths is taken. Find the probability that an individual man's step length is less than 1.9 feet. Round your answer to 4 decimal places, if necessary. Find the probability that the mean of the sample taken is less than 1.9 feet. Round your answer to 4 decimal places, if necessary.
2nd VARS normalCDF lower: -1E99 upper: 1.9 μ= 2.5 σ= 0.5 =.115 lower: -1E99 upper: 1.9 μ= 2.5 σ= 0.5/√51 =0
The cost of 5 gallons of ice cream has a variance of 49 with a mean of 36 dollars during the summer. What is the probability that the sample mean would be less than 37.5 dollars if a sample of 142 5-gallon pails is randomly selected? Round your answer to four decimal places.
2nd VARS normalCDF lower: -1E99 upper: 37.5 μ= 36 σ= 7/√142 =.9947
Intelligence Quotient (IQ) scores are often reported to be normally distributed with μ=100.0 and σ=15.0 A random sample of 35 people is taken. What is the probability of a random person on the street having an IQ score of less than 95? Round your answer to 4 decimal places, if necessary.
2nd VARS normalCDF lower: -1E99 upper: 95 μ= 100 σ= 15 =.3694 lower: -1E99 upper: 95 μ= 100 σ= 15/√35 =.0243
Goofy's fast food center wishes to estimate the proportion of people in its city that will purchase its products. Suppose the true proportion is 0.06 If 269 are sampled, what is the probability that the sample proportion will differ from the population proportion by less than 0.04? Round your answer to four decimal places.
2nd VARS normalCDF lower: .06-.04 upper: .06+.04 μ= .06 σ= √.06(1-.06)/269 =
Suppose 54% of the population has a retirement account. If a random sample of size 555 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 5%? Round your answer to four decimal places.
2nd VARS normalCDF lower: .54-.05 upper: .54+.05 μ= .54 σ= √.54(1-.54)/555 = .9819
Suppose 50% of the doctors in a hospital are surgeons. If a sample of 576 doctors is selected, what is the probability that the sample proportion of surgeons will be greater than 55%? Round your answer to four decimal places.
2nd VARS normalCDF lower: .55 upper: 1E99 μ= .5 σ= √.5(1-.5)/57 =.0082
Suppose babies born in a large hospital have a mean weight of 4022 grams, and a standard deviation of 266 grams. If 53 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by less than 45 grams? Round your answer to four decimal places.
2nd VARS normalCDF lower: 4022-25 upper: 4022+25 μ= 4022 σ= 266/√53 =.78
A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places.
2nd VARS normalCDF lower: 533.2 upper: 1E99 μ= 530 σ= 58/√75 =.3164
Find the value of z such that 0.08 of the area lies to the right of z. Round your answer to two decimal places.
2nd, VARS, 3:invNorm area: 1-.08 μ= 0 σ= 1 =1.41
The life of light bulbs is distributed normally. The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours. Find the probability of a bulb lasting for at most 554 hours. Round your answer to four decimal places.
2nd, VARS, NormalCDF Lower: -1E99 Upper: 554 μ= 520 σ= 15 =.9883
Find the area under the standard normal curve between z=−1.76 and z=0.07. Round your answer to four decimal places, if necessary.
2nd, VARS, NormalCDF Lower: -1.76 Upper: 0.07 μ= 0 σ= 1 =.4887
Find the area under the standard normal curve to the LEFT of z=1.87. Round your answer to four decimal places, if necessary.
2nd, VARS, NormalCDF Lower: -1E99 Upper: 1.87 μ=0 σ=1 =.9693
probability distribution
a table or formula that gives the probabilities for every random variable where 1. must be between 0 and 1, 2. sums to 1
Calculate the standard score of the given X value, X=66.5, where μ=73.3 and σ=75.6. Round your answer to two decimal places.
66.5-73.3/75.6 = -.0899
random variable
a variable that is random (decimal places)
Find the standard deviation of the sampling distribution of sample means using the given information. Round to one decimal place, if necessary. μ=47 σ=9 n=4
9/√4 = 4.5
A film distribution manager calculates that 8% of the films released are flops. If the manager is correct, what is the probability that the proportion of flops in a sample of 564 released films would differ from the population proportion by greater than 3%? Round your answer to four decimal places.
?? .0086
What is the difference between a discrete random variable and a continuous random variable?
A discrete random variable takes a set of separate values (with gaps), while a continuous random variable has possible values that form an interval (no gaps)
A philosophy professor assigns letter grades on a test according to the following scheme. A: Top 12% of scores B: Scores below the top 12% and above the bottom 57% C: Scores below the top 43% and above the bottom 19% D: Scores below the top 81%81% and above the bottom 5% F: Bottom 5% of scores Scores on the test are normally distributed with a mean of 66.5 and a standard deviation of 9.9 Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.
ASK
The process of manufacturing a ball bearing results in weights that have an approximately normal A) Suppose you select one ball bearing at random, what is the probability that it weighs 0.148g? B) Suppose you select one ball bearing at random, what is the probability that it weighs less than 0.148g? C) What is the 95th percentile of weights of ball bearings? (That is, what weight will have 95% of the distribution below it?) D) Find the two numbers such that the middle 90% of ball bearings will have weights between them. E) Suppose you select a sample of 100 ball bearings at random, what does the central limit theorem tell us about the sampling distribution of the sample mean (x bar), of these 100 bearings? What is the probability that the sample mean is less than 0.148 kg?
A) 0 because the probability of being exactly equal to a specific value is 0 B) normalcdf lower: -1E99 upper: 0.148 μ: 0.15 σ: .003 =.2525 So, it is a 25% chance! C) invnorm area: 1-.95 μ: 0.15 σ: .003 =.1451 D) The central 90% would be between the 5th and 95th percentiles, so call them L and R L = 5th percentile invnorm area: .05 μ: 0.15 σ: .003 =.1451 R= 95th percentile invnorm area: .95 μ: 0.15 σ: .003 =.1549 E) normal cdf lower: -1E99 upper: .148 μ: 0.15 σ: .003/√100 = 1.315 E 10^-11 It's super small, so it probably will not happen.
In the general US population, it has been reported that 25% of adults smoke. In a recent survey of college students, 12.9% of the 505 subjects surveyed said they smoked. A) Construct a 99% confidence interval for the true population proportion of college students who smoke. Provide the correct interpretation of the interval. B) Does the interval provide sufficient evidence to conclude the proportion of college students who smoke is different than .25? Explain why or why not. C) What was the margin of error for interval in part a? D) Suppose some researchers want to repeat this study in a similar population at another university, but they require a margin of error of only 4%. What size sample would they need to have?
A) 1 Prop Zint .129(505) = 65,145 (round to 65) n=505 the interval is (.09033,.1671) - We are 99% confidence that the true proportion of smokers in our population is in this interval B) Yes, because the interval excludes .25, so we can be 99% confident the population proportion is NOT .25. C) .1671-.1287 = .0384 D) 2.576^2(.1287)(1-.1287)/.042 =465.07 We need 466 subjects
In a local factory, they produce breakfast cereals. The printed net weight on the boxes is 18 oz. The actual weights of individual boxes are reasonably normally distributed, and the company wants the true mean weight to be slightly larger than 18, so they can have a high level of confidence that any randomly selected box will have at least 18 ounces in it (otherwise they could be subject to penalties). They take a random sample of 20 boxes from the line on a given day, and compute the sample mean to be 18.6 ounces and the sample standard deviation to be 2 ounces. A) Construct a 90% confidence interval for the true mean of all boxes produced. Provide the correct interpretation of the interval. B) From the t-table provided, or the calculator, what is the t-critical used in the above interval, and what is the margin of error? C) Does the interval provide sufficient evidence to conclude the mean of all boxes produced is greater than 18? Explain why or why not.
A) Tinterval x bar= 18.6 Sx= 2 (17.827,19.373) - We are 90% confident that the true mean of all boxes is in this interval B) 1.729(2)/√20 = .773 C) No, because the interval contains 18 (and values less than 18) so 18 is a plausible vale for the population mean.
At a university, 60% of the 7,400 students are female. The student newspaper takes a random sample of 50 students and surveys them. They report that 25 of the 50 in the sample were female. A) Find the mean and standard error of the sampling distribution of the sample proportion of females, for a sample of n=50. B) Is it "unusual" to get a proportion of .50 females or less in a sample of size 50, from a population that is .60 female?
A) mean = .6 standard error = √.6(1-.6)/50 =.0693 B) No, it's not unusual because .50 is less than 2 SE's away from .6. normalcdf lower: -1E99 upper: 0.5 μ: 0.6 σ: 0.0693 =0.745 Typically, a value isn't "unusual" unless there is less than a 5% (.05) chance of it occurring.
Serum cholesterol is an important risk factor for coronary disease. The level of serum cholesterol is normally distributed with a mean of 219 mg/dL and a standard deviation of 50 mg/dL. The clinically desirable range for serum cholesterol is < 200 mg/dL and serum cholesterol levels of over 250 mg/dL indicate a high-enough risk for heart disease to warrant treatment. A) What is the probability that a randomly selected person will have a borderline high serum cholesterol level (that is, > 200, but < 250 mg/dL? B) What is the probability that the sample mean from a randomly selected sample of 36 people will be > 200, but < 250 mg/dL?
A) normal cdf lower: 200 upper: 250 μ: 219 σ: 50 =.3804 B) normal cdf lower: 200 upper: 250 μ: 219 σ: 50/√36 =.9886
Let X denote the response of a randomly selected person to the question, "What is the ideal number of children for a family to have?" The probability distribution for X for females in the US is as shown below: X P(X) 0 0 1 0.03 2 0.63 3 0.23 4 ?? A) Fill in the missing probability B) Compute the mean of this distribution
A) the whole right side needs to add up to 1 .03+.63+.23 = .89 1-.89= .11 B) STAT, edit, enter numbers into L1 and L1, calc, 1 var stats mean= 2.42
Determine whether or not the given procedure results in a binomial distribution. If not, identify which condition is not met. Surveying 66 people to determine which brand of soft drink is their favorite.
First, ask "Are there only 2 outcomes of this experiment?" - No NOT a binomial distribution because there are more than 2 possible outcomes
A consumer affairs investigator records the repair cost for 27 randomly selected VCRs. A sample mean of $86.26 and standard deviation of $11.77 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Construct the 90% confidence interval. Round your answer to two decimal places.
First do: 1-confidence interval/2 (.1/2=.05) Next: do: n-1 (27-1=26) 2nd, VARS, invT area: .05 df: 26 =1.706 STAT, tests, Tinterval = (82.397, 90.123)
Determine whether or not the distribution is a discrete probability distribution and select the reason why or why not. x: -1, 1, 8 P(X=x): 0.34, 0.29, 0.37
Does the sum equal 1? 0.34, 0.29, 0.37 = 1 YES Are the values between 0 and 1, inclusively? YES so....... Yes, since the probabilities lie inclusively between 0 and 1, and the sum of the probabilities is equal to 1
T/F: The normal random variable X is the number of success in n trials
False - such a random variable would be binomial, not normal
Construct the discrete probability distribution for the random variable described. Express the probabilities as simplified fractions. The number of tails in 3 tosses of a coin. x 0 1 2 3 P(X=x) ________ __________ _________ ___________
First, find the total number possible outcomes 2^3= 8 (WRITE OUT ALL POSSIBLE OUTCOMES) HHH, HTH, HHT, THH TTTT, THT, TTH, HTT Determine the probability of 0 heads P(X=0) = 1/8 Determine the probability of 1 head P(X=1) = 3/8 Determine the probability of 2 head P(X=2) = 3/8 Determine the probability of 3 head P(X=3) = 1/8
A sociology professor assigns letter grades on a test according to the following scheme. A: Top 9% of scores B: Scores below the top 9% and above the bottom 63% C: Scores below the top 37% and above the bottom 24% D: Scores below the top 76% and above the bottom 7% F: Bottom 7% of scores Scores on the test are normally distributed with a mean of 73.9 and a standard deviation of 9.9 Find the numerical limits for a D grade. Round your answers to the nearest whole number, if necessary.
First, find upper limit: 100-76= 24 2nd, VARS, 3:invNorm area: .97 μ= 73.9 σ= 9.9 =67 Then, find lower limit: .07 area: .07 μ= 73.9 σ= 9.9 =59 (67,59)
NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier study, the population proportion was estimated to be 0.46. How large a sample would be required in order to estimate the fraction of people who black out at 66 or more Gs at the 95% confidence level with an error of at most 0.03? Round your answer up to the next integer.
Hit fraction button (right side of confidence interval chart)^2(pop. proportion)(1-pop proportion)/ (at most number)^2 (1.96)^2(.46)(1-.46)/(.03)^2 = 1060
A quality control inspector has drawn a sample of 18 light bulbs from a recent production lot. If the number of defective bulbs is 1 or less, the lot passes inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will pass inspection? Round your answer to four decimal places.
It's asking for P(X>1) (this is binomCDF) 2nd, VARS, binomcdf trials= 18 p= .3 x value= 1 = .0142
Which has higher payout?
LOOK AT 5.1 HOMEWORK
Which has least amount of risk?
LOOK AT 5.1 HOMEWORK
If you draw a card with a value of 4 or less from a standard deck of cards, I will pay you $18. If not, you pay me $9. (Aces are considered the highest card in the deck.) a) Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values. b) If you played 626 times, how much would you expect to win or lose?
LOOK AT NOTES FOR CHEAT SHEET OF _______ OR LESS
A research company desires to know the mean consumption of milk per week among males over age 45. They believe that the milk consumption has a mean of 4.7 liters, and want to construct a 99% confidence interval with a maximum error of 0.09 liters. Assuming a standard deviation of 1.51 liters, what is the minimum number of males over age 45 they must include in their sample? Round your answer up to the next integer.
Look at level of confidence chart for 99% (2.575) (2.575 x 1.5/.09)^2 =1840
A soft drink manufacturer wishes to know how many soft drinks teenagers drink each week. They want to construct a 90% confidence interval with an error of no more than 0.08. A consultant has informed them that a previous study found the mean to be 5 soft drinks per week and found the standard deviation to be 1. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.
Look at level of confidence for 90% (1.645) (1.645 x 1/.08)^2 =433
The mean amount of money spent on lunch per week for a sample of 233 college students is $43. If the margin of error for the population mean with a 98% confidence interval is 2.50, construct a 98% confidence interval for the mean amount of money spent on lunch per week for all college students.
Lower Endpoint: 43-2.5 = 40.5 Upper Endpoint: 43+2.5= 45.5
A marketing research company desires to know the mean consumption of milk per week among males over age 32. A sample of 710 males over age 32 was drawn and the mean milk consumption was 4.6 liters. Assume that the population standard deviation is known to be 0.8 liters. Construct the 98% confidence interval for the mean consumption of milk among males over age 32. Round your answers to one decimal place.
Lower endpoint: mean - right side of confidence interval x (stdev/√n) Upper endpoint: mean + right side of confidence interval x (stdev/√n) lower: 4.6-2.33(0.8/√710) = 4.5 upper: 4.6+2.33(0.8/√710) = 4.7
Can a normal approximation be used for a sampling distribution of sample means from a population with μ=39 σ=8 n=4?
No, because the sample size is less than 30
Determine whether or not the given procedure results in a binomial distribution. If not, identify which condition is not met. Surveying 81 people to determine which statistics course they have taken.
No, because there aren't only 2 outcomes
In the previous question, after the student determines the 95% confidence interval, they conclude that 95% of Kansas students had a score between the endpoints of the interval. Is this a correct interpretation of the confidence interval? Why or why not?
No, not correct because a confidence interval gives information about the MEAN of population, not individual students. - They should say, "We are 95% confident the true population mean is the interval"
An environmentalist wants to find out the fraction of oil tankers that have spills each month. Suppose a sample of 836 tankers is drawn. Of these ships, 586 did not have spills. Using the data, estimate the proportion of oil tankers that had spills. Enter your answer as a fraction or a decimal number rounded to three decimal places.term-111 Suppose a sample of 836 tankers is drawn. Of these ships, 586 did not have spills. Using the data, construct the 99% confidence interval for the population proportion of oil tankers that have spills each month. Round your answers to three decimal places.
Read question carefully. It's asking who HAD spills but it gives you who did not have spills. So, do 836-586 = 250. Now take 250/836 = .299 ********STAT, test, 1-PropZint x=250 n=836 c-level= .99 =(.25826, .33983)
discrete probability distribution
consists of the values a random variable can assume and the corresponding probabilities of the values
The temperature in Celsius in an Italian city Discrete or Continous?
continuous
binomcdf
cumulative binomial probability
How will decreasing the sample size without changing the level of confidence affect the width of a confidence interval for a population mean? Assume that the population standard deviation is unknown and the population distribution is approximately normal.
The margin of error will increase because the critical value will increase, and the sample size is located in the denominator of the formula for margin of error. The increased margin of error will cause the confidence interval to be wider
What happens to the width of a confidence interval for a population mean if the level of confidence is increased without changing the sample size? Assume that the population standard deviation is unknown and the population distribution is approximately normal.
The margin of error will increase because the critical value will increase. The increased margin of error will cause the confidence interval to be wider
The number of people who quit smoking in each of the last ten years. Discrete or Continuous?
discrete
Determine whether or not the given procedure results in a binomial distribution. If not, identify which condition is not met. Surveying 81 people to determine if they have taken a statistics course.
Yes, there are only 2 outcomes
normal distribution
a continuous probability distribution for a continuous random variable, X, defined completely by it's mean and standard deviation, if the following properties are true: 1. a normal distribution is bell-shaped and symmetric abouts its mean 2. A normal distribution is completely defined by its mean and standard deviation 3. The total area under a normal distribution curve equals 1 4. The x-axis is a horizontal asymptote for a normal distribution curve
continuous random variable
a continuous variable whose numeric value is determined by the outcome of a probability experiment
Consider the value of t such that the area under the curve between −|t| and |t| equals 0.95 Assuming the degrees of freedom equals 8, select the t value from the t table.
Use degree of freedom chart 1-0.95=.05 .05/2 = .025 Now go down to where it says 8 on the left side and .025 across the top.
normal distribution function
a normal distribution with a mean equal to 0 and a standard deviation equal to 1
unbiased estimate
a point estimate that does not consistently underestimate or overestimate the population parameter
Consider the following data: x 1 2 3 4 5 P(X=x) 0.1 0.2 0.3 0.2 0.2 a) Find the expected value E(X). Round your answer to one decimal place. b) Find the variance c) Find the standard deviation d) Find the value of P(X ≥ 5). e) Find the value of P(X ≤ 3).
a) E(X) = 3.2 hit STAT, 1: Edit, type in numbers into L1 and L2, then hit STAT again, scroll right to calc, 1: 1-Var Stats, List: L1, FreqList: L2, enter. b) 1.25^2 = 1.5625 Look at standard deviation (σ) and square it to find the variance c) 1.25 Look at σ d) 0.2 Add everything that is greater than 5 e) 0.6 Add everything that is below 3
T/F: the mean of a random variable must always be a possible value of the random variable
false
True or False? A sampling distribution of sample means has a mean equal to the population mean, μ, divided by the sample size.
false
confidence interval
an interval estimate associated with a certain level of confidence
The online shopping industry is growing at an extraordinary rate. Current estimates suggest that 20% of people in the US with home-based computers regularly purchase items online. Suppose that 15 people with home-based computers were randomly and independently sampled. What is the probability that at least 1 of those sampled will say YES, they regularly purchase items online? 0.0352 0.8328 0.9648 0.1671 0.1319
binompdf: trials: 15 p: .2 x-value: .0 =.0352 1-.0352= .9640
central limit theorem (clt)
for any given population with a mean and standard deviation, a sampling distribution of sample means will have the following 3 characteristics if either the sample size is at lead 30 or the population is normally distributed 1. the mean of a sampling distribution of sample means 2. the standard deviation of a sampling distribution of samples means is standard error 3. the shape of a sampling distribution of sample means will approach that of a normal distribution, regardless of the shape of the population sample
Find the standard deviation of the following data x -6 -5 -4 -3 -2 -1 P(X=x) 0.2 0.1 0.2 0.2 0.1 0.2
hit STAT, 1: Edit, type in numbers into L1 and L2, then hit STAT again, scroll right to calc, 1: 1-Var Stats σ = 1.7
inflection point
indicates a point on a curve where the curve changes from curving up to curving down, or vice versa
critical z-values
mark the boundaries for the area under the middle of the standard normal curve that corresponds with a particular level of confidence
binomial distribution
the discrete probability distribution with a fixed number of independent trials, where each trial has only 2 possible outcomes and one of these outcomes is counted: 1. Experiment has a fixed number of identical trials 2. each trial is independent 3. only 2 outcomes for each trial 4. probability of getting a success = P probability of getting a failure = 1-P 5. random variable (x) counts the number of successes in trials (n)
sampling distribution
the distribution of all the values of a particular sample statistic for ALL possible samples of a given size, n
population distribution
the distribution of the values of a particular sample statistic for all possible samples of a given size, n
sampling distribution of sample means
the distributions of sample means for all possible samples of a given size, n
population parameter
the fraction or percentage of a population that displays a certain characteristics
sample propportion
the fractional part of a sample that displays a certain characteristic
margin of error/maximum error of estimate
the largest possible distance from the point estimate that a confidence interval will cover
level of confidence
the percentage of all possible samples of a given size that will produce interval estimates that contain the actual parameter
An exit poll in a recent election was conducted in order to predict the winner of the evening news, prior to the polls officially closing. If 1000 people were randomly selected as they left the polls, and 53% say they voted for Cleetus Aardvark. If the truth in the population is that it is a tie (i.e. 50% support Cleetus), what is the probability that a poll of 1000 people could have a proportion of .53 or higher? Would you be willing to declare him the winner?
√0.5(1-0.5)/1000 = .0158 normal cdf lower: .53 upper: 1E99 μ: .5 σ: .0158 =.0289
