round all answers to 4 decimal places where possible.

Question

The average amount of money spent for lunch per person in the college cafeteria is $6.12 and the standard deviation is $2.18. Suppose that 48 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.

1. What is the distribution of ?  ~ N(Correct,)
2. What is the distribution of ?  ~ N(Correct,)
3. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $6.29 and $6.47. 
4. For the group of 48 patrons, find the probability that the average lunch cost is between $6.29 and $6.47. 
5. For part d), is the assumption that the distribution is normal necessary? 

Answer 1

x is N(6.12,2.18)

xbar is N (6.12, 2.18/sqrt(48)) = N(6.12,0.315)

want to find P(6.29 < x < 6.47).

z statistic = (x - mean)/SD is N(0,1)

apply transformation to x:

P(6.29 < x < 6.47) = P( (6.29- 6.12)/2.18 < z < (6.47 - 6.12)/2.18) = P(0.08 < z < 0.16) = P(z < 0.16) - P(z < 0.08) = 0.5636 - 0.5319 = 0.0317

want to find P(6.29 < xbar < 6.47)

z-statistic = (xbar - mean)/SE, where SE = standard error = 0.0315

apply transformation to xbar:

P(6.29 < xbar < 6.47) = P( (6.29- 6.12)/0.315 < z < (6.47 - 6.12)/0.315) = P(0.54 < z < 1.11) = P(z < 1/11) - P(z < 0.54) = 0.8665 - 0.8054 = 0.0611

For sample sizes greater than 30, the distribution of the sample means is normal, so the assumption of normality is not necessary for the previous problem.