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Exploring Sampling Distributions and the Central Limit Theorem Using Minitab Lab 2

Explores sampling distributions and the Central Limit Theorem using Minitab software.

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    Page 1 of 7 Lab 2 Exploring Sampling Distributions and the Central Limit Theorem Using Minitab Lab 2 Objectives:  Learn to use Minitab to compute probabilities and percentile of distributions .  Use simulation to understand central limit theorem and sampling distributions  1. Using Minitab, answer the following. Write your answers in your document that you will submit. Your answer must be complete. Remember that Minitab only helps you to calculate your answer. You need to provide all the details such as proper definitions, proper set - up of the problem and explanation of the process. You will not get much credit without all this information. The time required for Speedy Lube to complete an oil change service on an automobile follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2.5 min. Question 1a: Speedy Lube guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half - price. What percent of customers rec eive the service for half - price? Solution: Let X be the random variable then the probability will be P(X>20)=P( 𝑋 − 1଻ 2 . 5 > 20 − 1଻ 2 . 5 ) = 𝑃 ( 𝑍 > 1 . 2 ) = 0 . 1151 where Z= 𝑋 − 1଻ 2 . 5 Hence 11.51% of customer receives the service for half price . Question 1b : If Speedy Lube does not want to give the discount to more than 3% of its cu s tomers, how long should it make the guaranteed time limit? Solution: P(Z ≥ 𝑍 0.03 ) ≤ 0 . 03  P( 𝑋 − 1଻ 2 . 5 ≥ 1 . 181 ) ≤ 0 . 03  P(X ≥ 19 . 9525 ) ≤ 0 . 03  Hence the granted limit is 19.9525.  Question 1c : What fraction of autos take between 11 min. and 16 min.? Solution: P(11 ≤ 𝑋 ≤ 16 ) = 𝑃 ( 11 − 1଻ 2 . 5 ≤ 𝑋 − 1଻ 2 . 5 ≤ 16 − 1଻ 2 . 5 ) = 𝑃 ( − 2 . 4 ≤ 𝑍 ≤ − 0 . 4 ) = 0 . 3364 2. Next we see the idea of using simulation to generate a sample from a population. The population distribution is specified and we draw a sample from this population by generating random numbers from the population distribution. We will learn about the behavior of the statistics computed for several samples from the population. The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the same population. Recall that a statistic is a measure of a sample, whereas a parameter is a measure of a population.

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    Page 2 of 7 Lab 2 Suppose the number of toys a child plays with in a child care center on a given day is a random variable. Suppose the probability distribution of X is given by x 0 1 2 3 4 5 P(X=x) 0.03 0.06 0.1 0.13 0.47 0.21 i. Choose Calc -- > Setbase and enter the l your student identity number (not your social security number) . Click “OK”. ii. Enter the values 0,1,2,3,4,5 in column c1. Name this column x . iii. Enter the values 0.03, 0.06, 0.1, 0.13, 0.47, 0.21 in column c2. Name this column P(X=x) . So, the pmf of X is given in columns C1 and C2. Let us consider this data , the number of toys a child plays with, as the population. So, the population contains values 0,1,2,3,4,5 in the relative frequency given in C2. Columns C1 and C2 provide the population distribution. iv. Graph this population distribution using Graph -- > Scatterplot, selecting the Simple graph option; choose the column with values of the random variable as the x variable and the column with the probabilities as the y variable. Click the Dataview button and in the Dataview menu, select only project lines. Copy and paste(special) the graph in your document. The graph is as follows: Question 2a. : Comment on the shape of the ( population ) distribution that is represented in your graph. 5 4 3 2 1 0 0.5 0.4 0.3 0.2 0.1 0.0 x P ( X = x ) Scatterplot of P(X=x) vs x

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