M5 Problem Set

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Statistics

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Jan 9, 2024

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M5: Problem Set Due No due date Points 5 Questions 7 Time Limit None Attempt History Attempt Time Score LATEST Attempt 1 199 minutes 5outof 5 Score for this quiz: S out of 5 Submitted Aug 22 at 4:26pm This attempt took 199 minutes. Question 1 0/0 pts Suppose that you are attempting to estimate the weight of 600 parts. In order to use the infinite standard deviation formula, what sample size, n, should you use? Your Answer: n < (.05) 600 n < 30 Sample size must be less than 30
In order to use infinite standard deviation formula, we must have: n < 0.05(600) n <30 So, the sample size must be less than 30. Question 2 0/0 pts Suppose that you take a sample of size 15 from a population that is known to be normally distributed. Can the sampling distribution of X be approximated by a normal probability distribution? Your Answer: Yes, the sampling distribution of X can be approximated by a normal probability distribution because the population is normally distributed. Yes, the sampling distribution of X can be approximated by a normal probability distribution because the population is normally distributed. (Recall that if the population is normally distributed, the sampling distribution of X can be approximated by a normal probability distribution even for very small sample sizes.) Question 3 0/0 pts Suppose that you take a sample of size 20 from a population that is not normally distributed. Can the sampling distribution of X be approximated
by a normal probability distribution? Your Answer: No, the sampling distribution of X cannot be approximated by a normal probability distribution in this case. No, the sampling distribution of X cannot be approximated by a normal probability distribution in this case. (Recall that if the ~ population is not normally distributed, the sample size must be at ' least 30.) Question 4 0/0 pts Suppose a pharmaceutical company wants to do a study of the commissions of its sales force. Let's assume that there are 4,300 sales people and the population mean for the sales force is $52,400 in commissions and has a population standard deviation of $3,500. What is the probability that a simple random sample of 50 members of the sales force will have commissions within $400 of the population mean? Your Answer: Standard deviation of the sample distribution ‘05 o _ 3500 _ 494 97 vn /B0 Z-score _ ZT—p __ 52000-52400 _ —400 _ ~ 2= = T prer - = migr 3081~ '808‘ and _ T—p 5280052400 _ —400 ~ _ = oz T T 40497 19497 8081 ~ '808‘ Round values to -.81 and .81 |P(—.81<Z < .81)=P(Z< .81)— P(Z— .81) =.79103 .2
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There is a 0.58206 probability that a simple random sample of 50 members of the sales force will provide a sample mean that is within the $400 of the sample mean. and There is a 1 - .58206= .41794 probability the sample mean will not be within the $400 range.
We calculate the standard deviation of the sample distribution: o 3500 494,97 O = = —= 5 Wm0 We now have the necessary information needed to determine the - probability the sample mean x will be between $52,000 and - $52,800 this is the range which is within $400 of the population mean p of $52,400.) Calculate the z-score: X—pu Oz Zo= Using this formula, we will calculate the two z-scores that we will use to answer our question. _¥—p _ 52,000—52400 —400 2= T 49497 49497 —0.8081 ~ —.808 and _X—p _52800-52400 400 T ooy 494.97 "~ 494,97 z = 0.8081 ~ .808 We will round these values to -.81 and .81, respectively. So, we want to find P(-.81< Z < .81) on the standard normal probability distribution table. Recall that P(-.81< Z < .81)=P(Z<.81)-P(Z<-.81)=.79103 -.20897 = .582086. So now we know that there is a 0.58206 probability that a simple random sample of 50 members of the sales force will provide a sample mean that is within the $400 of the sample mean. Conversely, there is a 1 - .58206= .41794 probability the sample mean will not be within the $400 range.
Question 5 0/0 pts Suppose that you have a large number of potatoes. The mean weight of the potatoes is 7.4 ounces with a standard deviation of 2.2 ounces. We may assume a normal distribution. a) If you choose 25 potatoes at random, what is the probability that the mean of this sample of 25 potatoes is between 7 and 8 ounces? b) If you choose 45 potatoes at random, what is the probability that the mean of this sample of 45 potatoes is less than 7 ounces? c) If you choose 40 potatoes at random, what is the probability that the mean of this sample of 40 potatoes is more than 7.5 ounces? Your Answer: a) Standard deviation o 2.2 or = = == = 44 Vn V25 z-score for 7 oz _T—p_ 7T-74 zZ = =@ 91 z-score for 8 oz _zT—p 874 _ 2= = = 1.36 P(-91< Z < 1.36) P(-91< Z < 1.36)=P(Z<1.36)-P(Z<-.91)=.91309-.18141=.73168 There is a .73168 probability that a random sample of 25 potatoes with have a mean weight between 7 and 8 ounces.
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b) Standard deviation o 2.2 or = = == = 44 \/fi V25 Z-score TR _T-T4 _ {99 2 = = oT 328 Probability distribution P(Z <-1.22) = 11123 There is a .11123 probability that a simple random sample of 45 potatoes will have a mean weight less than 7 ounces. c) Standard deviation o 2.2 or = = == = 34785 Vn 40 Z-score _zTz—p _ 75-74 2= Tz 347851 29 P(Z > .29) = 1— .61409 = .38591
a) We calculate the standard deviation of the sample distribution: ,__22_ *Tyn V25 So, we want to find P(-.91< Z < 1.36) on the standard normal ~ probability distribution table. Recall that - P(-.91< Z < 1.36)=P(Z<1.36)-P(Z<-.91)=.91309-.18141=.73168. Therefore, there is a .73168 probability that a random sample of 25 potatoes with have a mean weight between 7 and 8 ounces. b) We calculate the standard deviation of the sample distribution: 2.2 0= =—=——=.,3280 Y oyn 45 Calculate the z-score: _X-p_7-74_ ., 2= T 3280 So, we want to find P(Z < -1.22) on the standard normal probability distribution table. P(Z <-1.22) = 11123. Therefore, there is a .11123 probability that a simple random sample of 45 potatoes will have a mean weight less than 7 ounces. c) We calculate the standard deviation of the sample distribution: 7~ _ 0V 0V o
Ldalculatle ine Z-score. So, we want to find P(Z > .29) on the standard normal probability distribution table. Recall that P(Z > .29) = 1 - P(Z < .29) = 1-.61409=.38591. Therefore, there is a .38591 probability that a simple random sample of 40 potatoes will have a mean weight greater than 7.5 ounces. Question 6 0/0 pts According to the American Diabetes Association (2013), 8.3 % of the U.S. population have diabetes. a) Suppose that you take a random of 80 Americans, what is the probability that 7 % or less of these 80 people have diabetes? b) Suppose that you take a random of 120 Americans, what is the probability that 8 % or more of these 120 people have diabetes? Your Answer: a.) p=.083 n=80 o \/p(ln—p) _ \/.083(;.083) _ .030845| __ p—p __ .07-.083 2= o5 T 030845 _'42| |P(Z < —.42) .33724
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n=120 o p(l_n—p) _ \/.083(;2—0.083) 095184 __ p—p __ .07-.083 ?= T = Tomet 12 —Pp P P(Z > —.12) =1 .45224 = 54776
o 0.083(1 0.083 o5 = Ju = J ( ) _ 0.03084 n 80 Now we use our old friend the z-score to determine the probability of our sample size yielding a proportional mean p within the desired range: p—p 0.07-0.083 oz 0.03084 b Z = We want P(Z<-0.42). From the standard normal table, we find: P(Z<-0.42)=.33724. So there is a .33724 probability that the percentage of the sample that has diabetes is less than 7 %. b) ~ ,p(l —p) _ /0.083(1—0.083) 05 = ——= j = = 0.0252 Now we use our old friend the z-score to determine the probability of our sample size yielding a proportional mean p within the desired range: p—p 0.08—0.083 PR Z. = We want P(Z>-0.12). From the standard normal table, we find: P(Z>-.12)=1- P(Z<-.12)=1-.45224=.54776. So there is a .54776.probability that the percentage of the sample that has diabetes is greater than 8 %.
Question 7 5/5pts As a reminder, the questions in this review quiz are a requirement of the course and the best way to prepare for the module exam. Did you complete all questions in their entirety and show your work? Your Answer: Yes Quiz Score: 5 out of 5
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