The California Chamber of Commerce has asked you to do a study concerning people
ID: 3172882 • Letter: T
Question
The California Chamber of Commerce has asked you to do a study concerning people's attitudes toward our city. As part of the study, you will ask them to rate their image of California on a scale from 1 to 100 (1= awful city - call in the bulldozers; 100 = wonderful city - there is none better). It is known that this scale is normally distributed with an average rating of 50 and a standard deviation of 12.
a. What is the probability that one person chosen at random gives California a rating of:
(i) 65 or better?
(ii) 53 or worse?
(iii) somewhere between 50 and 55?
b. Suppose a sample of 64 people is chosen.
(i) What is the probability that the average of the sample is between 70 and 64?
(ii) Challenge: Construct a 95% confidence interval for the mean of this sample.
Explanation / Answer
Q1.
Mean ( u ) =50
Standard Deviation ( sd )=12
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
P(X < 65) = (65-50)/12
= 15/12= 1.25
= P ( Z <1.25) From Standard Normal Table
= 0.8944
P(X > = 65) = (1 - P(X < 65)
= 1 - 0.8944 = 0.1056
b.
P(X > 53) = (53-50)/12
= 3/12 = 0.25
= P ( Z >0.25) From Standard Normal Table
= 0.4013
P(X < = 53) = (1 - P(X > 53)
= 1 - 0.4013 = 0.5987
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 50) = (50-50)/12
= 0/12 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 55) = (55-50)/12
= 5/12 = 0.4167
= P ( Z <0.4167) From Standard Normal Table
= 0.66154
P(50 < X < 55) = 0.66154-0.5 = 0.1615
Q2.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 64) = (64-50)/12/ Sqrt ( 64 )
= 14/1.5
= 9.3333
= P ( Z <9.3333) From Standard Normal Table
= 1
P(X < 70) = (70-50)/12/ Sqrt ( 64 )
= 20/1.5 = 13.3333
= P ( Z <13.3333) From Standard Normal Table
= 1
P(64 < X < 70) = 1-1 = 0