Question

Sale amounts during lunch hour at a local subway are normally ditributed, with a mean $ 8.35, and a standard deviation og $ 1.1. a. Find the probability that a randomly selected sale was at least $ 9.42? b. A particular sale was $ 10.28. What is the percentile rank for this sale amount? % c. Give the sale amount that is the cutoff for the highest 72 %? d.What is the probability that a randomly selected sale is between $5.00 and $8.00? e. What sale amount represents the cutoff for the middle 34 percent of sales? ( The smaller number here)(Bigger number here)

Answer 1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answer.

Given mean = 8.35

S.d = 1.1

Z = (x-mean)/s.d

A)

P(x>9.42)

Z = (9.42 - 8.35)/(1.1)

Z = 0.97

From z table, P(Z>0.97) = 0.1660

B)

Z = (10.28 - 8.35)/1.1

Z = 1.75

From z table, P(Z<1.75) = 0.9599

C)

From z table, P(z>-0.58) = 0.72

Therefore, z = -0.58

Z = (x-mean)/s.d

-0.58 = (x-8.35)/1.1

X = 7.712

D)

P(5<x<8) = P(x<8) - p(x<5)

P(x<8)

Z = (8-8.35)/1.1

Z = -0.32

From z table, P(z<-0.32) = 0.3745

P(x<5)

Z = -3.05

From z table, P(Z<-3.05) = 0.0011

Required probability is = 0.3745 - 0.0011 = 0.3734

E)

First we need to determine the area in each tail

100-34 = 66

So, area in each tail = 66/2 = 33% = 0.33

From z table, P(z<-0.44) = 0.33

X = 8.35 - (1.1*0.44) = 7.866 (smaller number)

From z table, P(z>0.44) = 0.33

X = 8.35 + (1.1*0.44) = 8.834 (Bigger number)