Question

The average amount of money that people spend at McDonalds fast food place is $7.4000 with a standard deviation of $1.6700. 11 customers are randomly selected. Please answer the following questions, and round all answers to 4 decimal places where possible and assume a normal distribution.

A.)What is the distribution of X? X ~ N()

B.)What is the distribution of ¯x? ¯x ~ N()

C.)What is the distribution of ∑x? ∑x ~ N(,)

D.)What is the probability that one randomly selected customer will spend more than $7.4148?

E.)For the 11 customers, find the probability that their average spent is more than $7.4148.

F.)Find the probability that the randomly selected 11 customers will spend more than $81.5628.

G.)For part e) and f), is the assumption of normal necessary? No or Yes

H.)The owner of McDonalds gives a coupon for a free sundae to the 4% of all groups of 11 people who spend the most money. At least how much must a group of 11 spend in total to get the free sundae? $

Answer 1

Answer:

Given,

Mean = 7.4

Standard deviation = 1.67

Sample = 11

a)

X ~ N(7.4 , 1.67)

b)

Xbar ~ N(7.4 , 1.67/sqrt(11)) ~ N(7.4 , 0.50)

c)

∑x ~ N(7.4*11 , sqrt(11)*1.67) = ∑x ~ N(81.4 , 5.54)

d)

P(X > 7.4148) = P(z > (7.4148 - 7.4)/1.67)

= P(z > 0.01)

= 0.4960106 [since from z table]

= 0.4960

e)

P(Xbar > 7.4148) = P(z > (7.4148 - 7.4)/0.50)

= P(z > 0.03)

= 0.4880335 [since from z table]

= 0.4880

f)

P(∑X > 81.5628) = P(z > (81.5628 - 81.4)/5.54)

= P(z > 0.03)

= 0.4880335 [since from z table]

= 0.4880

g)

Yes

h)

P(∑x > x) = 0.04

P(z > (x - 81.4)/5.54) = 0.04

since from z table

(x - 81.4)/5.54 = 1.755

x = 91.1227