Question

In: Statistics and Probability

In this assignment, you will learn how to describe the distribution of the sample mean (),...

In this assignment, you will learn how to describe the distribution of the sample mean (), and investigate how sample size affects variability of the sample mean (). To find out various probability values you will have to calculate the standard error of, and use the standard error to convert into Z score.

MARS Chocolate claims* that the net weight of its 31 oz Party Bucket is normally distributed with a mean of 32 oz and a standard deviation of 2 oz.

Assuming that the claim is correct, answer the following questions:

You must draw diagrams and show workings for questions (a) to (c).

(a) You purchase one of these Party Buckets, empty the content and measure its weight (denoted X). What is the chance that the weight (X) is less than 30 oz?

(b) Suppose you purchase 4 buckets and measure their mean weight ().

(i) Describe the probability distribution of the mean weight ().

(ii) What is the probability the mean weight () is less than 30 oz?

(c) Now suppose you purchase 9 buckets for a huge party and measure the mean weight ().

(i) Describe the probability distribution of the mean weight ().

(ii) What is the probability the mean weight () is less than 30 oz?

(d) Explain why your answer to (c)(ii) is different from (b)(ii).

Solutions

Expert Solution

a)

as z score =(X-mean)/std deviation

hence chance that the weight (X) is less than 30 oz =P(X<30)=P(Z<(30-32)/2)=P(Z<-1)=0.1587

b)

i)as population distribution is normal therefore sampling distribution of mean also is normal

probabiliy distribution of mean has

mean weight =32

and std error of mean =std deviaiton/sqrt(n)=2/sqrt(4)=1

probability the mean weight () is less than 30 oz =P(Xbar<30)=P(Z<(30-32)/1)=P(Z<-2)=0.0228

c)

probabiliy distribution of mean has

mean weight =32

and std error of mean =std deviaiton/sqrt(n)=2/sqrt(9)=0.67

probability the mean weight () is less than 30 oz =P(Xbar<30)=P(Z<(30-32)/0.67)=P(Z<-3)=0.0013

d)

fr we can see that as sample size increases ; std error of mean decreases so farther a sample mean from population mean lower the probability of such event for higher size sample ; and therefore probabiliy in part c is smaller from part b


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