Download COMSATS Institute of Information Technology Islamabad

COMSATS Institute of Information Technology Islamabad
MTH 161 Introduction to Statistics
Solution of Quiz 5
Program: BBA
Date: Dec 19, 2012
Total Points: 10
Instructor: Shahzad Fazal Bhatti
Solution for Left
The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours
and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period
of time
a. more than 19.5 hours?
Let X = time taken to assemble a randomly selected car, then X is a normal random variable with mean
20 hours and a standard deviation of 2 hours.
Suppose x1 = 19.5, then
x1 − 20
19.5 − 20
z1 =
=
= −0.25
2
2
P (X > 19.5) = 1 − P (X < 19.5)
= 1 − P (Z < −0.25) = 1 − 0.4013 = 0.5987
b. between 20 and 22 hours?
Suppose x2 = 20 and x3 = 22, then
z2 =
x1 − 20
20 − 20
=
= 0.00
2
2
z3 =
22 − 20
x1 − 20
=
= 1.00
2
2
P (20 < X < 22) = P (X < 22) − P (X < 20)
= P (Z < 1.00) − P (Z < 0.00)
= 0.8413 − 0.50 = 0.3413
Solution for Right
The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months
and standard deviation of 2 months. Find the probability that an instrument produced by this machine will last
a. less than 7 months.
Let Y = life span of a randomly selected instrument produced by the machine, then Y is a normal random
variable with mean 12 months and a standard deviation of 2 months.
Suppose y1 = 7, then
y1 − 12
7 − 12
z1 =
=
= −2.5
2
2
P (Y < 7) = P (Z < −2.5) = 0.0062
b. between 7 and 12 months.
suppose that y2 = 12, then
z2 =
y2 − 12
12 − 12
=
= 0.00
2
2
P (7 < X < 12) = P (X < 12) − P (X < 7)
= P (Z < 0.00) − P (Z < −2.5)
= 0.50 − 0.0062 = 0.4938
Points Distribution:
Part (a)
Part (b)
Total
4
6
10