Question

The average amount of money that people spend at Don Mcalds fast food place is $7.3800 with a standard deviation of $1.6900. 48 customers are randomly selected. Please answer the following questions, and round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X X ? X X ~ N(,) What is the distribution of ¯ x x¯ ? ¯ x x¯ ~ N(,) What is the distribution of ∑ x ∑x ? ∑ x ∑x ~ N(,) What is the probability that one randomly selected customer will spend more than $7.1642? For the 48 customers, find the probability that their average spent is more than $7.1642. Find the probability that the randomly s elected 48 customers will spend more than $343.8816. For part e) and f), is the assumption of normal necessary? NoYes The owner of Don Mcalds gives a coupon for a free sundae to the 2% of all groups of 48 people who spend the most money. At least how much must a group of 48 spend in total to get the free sundae? $

Answer 1

a)

x ~ N(7.38 , 1.69)

b)

x¯ ~ N(7.38 , 0.2439)

c)

∑x ~ N(354.24 , 11.7087)

d)

probability =P(X>7.1642)=P(Z>(7.1642-7.38)/1.69)=P(Z>-0.13)=1-P(Z<-0.13)=1-0.4492=0.5508

e)

probability =P(X>7.1642)=P(Z>(7.1642-7.38)/0.244)=P(Z>-0.88)=1-P(Z<-0.88)=1-0.1882=0.8118

f)

probability =P(X>343.8816)=P(Z>(343.8816-354.24)/11.709)=P(Z>-0.88)=1-P(Z<-0.88)=1-0.1882=0.8118

g)

No (since sample size >30)

h)

for 98th percentile critical value of z=		2.05
therefore corresponding value=mean+z*std deviation=			378.2867