Question

The average amount of money spent for lunch per person in the college cafeteria is $5.67 and the standard deviation is $2.76. Suppose that 45 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal placeswhere possible. What is the distribution of X ? X ~ N( , ) 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 $5.5729 and $5.8886. For the group of 45 patrons, find the probability that the average lunch cost is between $5.5729 and $5.8886. For part d), is the assumption that the distribution is normal necessary? NoYes

Answer 1

(a)

Distribution of X is Normal Distribution with mean = = 5.67 and SD = .

i.e.,

X N (5.67, 2.76)

(b)

SE = /

= 2.76/ = 0.4113

Distribution of is Normal Distribution with mean = = 5.67 and SD = 0.4113

i.e.,

X N (5.67, 0.4113)

(c)

To find P(5.5729< X < 5.8886)

Case 1: For X from 5.5729 to mid value:
Z = (5.5729 - 5.67)/2.76= - 0.0352

Table of Area Under Standard Normal Curve gives area = 0.0160

Case 2: For X from mid value to 5.8886:
Z = (5.8886 - 5.67)/2.76= 0.0792

Table of Area Under Standard Normal Curve gives area = 0.0319

So,

P(5.5729 < X < 5.8886) = 0.0160 + 0.0319 = 0.0479

So,

Answer is:

0.0479

(d)

To find P(5.5729< < 5.8886)

Case 1: For from 5.5729 to mid value:
Z = (5.5729 - 5.67)/0.4113 = - 0.2361

Table of Area Under Standard Normal Curve gives area = 0.0948

Case 2: For from mid value to 5.8886:
Z = (5.8886 - 5.67)/0.4113= 0.5315

Table of Area Under Standard Normal Curve gives area = 0.2019

So,

P(5.5729 < < 5.8886) = 0.0948 + 0.2019 = 0.2967

So,

Answer is:

0.2967

(e)

Correct option:

No

Explanation: Sample Size =n = 45 > 30 Large Sample. Central Limit Theorem applies.