Question

The average amount of money spent for lunch per person in the college cafeteria is $6.25 and the standard deviation is $2.28. Suppose that 17 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.What is the distribution of X ? X ~ N(6.25,2.28)What is the distribution of ¯x ? ¯x ~ N(,)For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $6.6005 and $6.927. For the group of 17 patrons, find the probability that the average lunch cost is between $6.6005 and $6.927. For part d), is the assumption that the distribution is normal necessary? NoYes

Answer 1

a)

We know that the distribution of the population is normally distributed with mean $6.25 and standard deviation $2.28 hence we have:

`X\sim N(6.25,2.28)`

b)

We know that the standard deviation of the mean is given by:

`\sigma_{\bar{x}}=(\sigma)/(√(n))`

Then in this case we have:

`\sigma_{\bar{x}}=(2.28)/(√(17))=0.553`

We also know that the mean of the distribution of the mean is equal to the mean of the population. Then the distribution of the mean is:

`X\sim N(6.25,0.553)`

c)

If we choose a random person we would need to use the distribution in part a. We need to find the probability:

`P(6.6005\leq X\leq6.927)`

to calculate this probability, we need to standardize the distribution, we achieve this by using the z-score defined as:

`z=(x-\mu)/(\sigma)`

Then we have:

`\begin{gathered} P(6.6005\leq X\leq6.927)=P((6.6005-6.25)/(2.28)\leq Z\leq(6.927-6.25)/(2.28)) \\ =P(0.1537\leq Z\leq0.2969) \\ =0.0557 \end{gathered}`

Therefore, the probability in this case is 0.0557

d)

In this case, since we are choosing from the group of 17 patrons, we will use the distribution on part b. Doing the same as the previous point we have:

`\begin{gathered} P(6.6005\leq X\leq6.927)=P((6.6005-6.25)/(0.553)\leq Z\leq(6.927-6.25)/(0.553)) \\ =P(0.6338\leq Z\leq1.2242) \\ =0.1527 \end{gathered}`

Therefore, the probability in this case is 0.1527

e)

Since the sample is less than 30, the assumption of normality is relevant, therefore, the answer is Yes