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Statistics_Normal distribution_Questions and answers

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This document includes model answers for 11 problems in Normal distribution analysis

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Statistics Questions and answers
Normal Distribution


1. A radar unit fixed in the road to measure speeds of cars. The speeds are found
to be normally distributed with a mean of 90 km/hr and a standard deviation of
10 km/hr. What is the probability that a car selected at random has a speed
more than 100 km/hr?

Answer
Assume x to be the random variable represents the speed of cars. We need to
find the probability that x is greater than 100, or P(x > 100)
For x = 100 , z = (100 - 90) / 10 = 1
P(x > 90) = P(z > 1) = 1 - 0.8413 = 0.1587

Therefore, probability that a car selected at a random has a speed more than 100
km/hr equals to 0.1587.


2. Most colleges of engineering require candidates for admission to take the
qualifire test. Scores of this test are generally normally distirbuted with a mean
of 527 and a standard deviation of 112. What is the likelihood of a singular
scoring over 500 to take place?



Answer
 1-0.4052=

 0.4052
0.5948

 500 527
Z  0.24107
112
= 527
= 112 500 527 X

-0.24 0 Z




1

, P X > 500 = P Z > -0.24 = 1 – 0.4052 = 0.594


3. What is the highest score must an individual score on this test in
order to score in the highest 5%?

Answer

= 527
= 112
P(X > ?) = 0.05 ⇒ P(Z > ?) =
0.05 P(Z < ?) = 1 - 0.05 = 0.95 ⇒
Z = 1.645
X = 527 + 1.645(112)
X = 527 + 184.24
X = 711.24



4. For a certain type of laptops, the length of time between charges of
the battery is normally distributed with a mean of 50 hours and a
standard deviation of 15 hours. Frank bought one of these laptops
and wants to know the probability that the length of time will be
between 50 and 70 hours.

Answer
If x the random variable representing the length of time with a mean of
50 and a standard deviation of 15. Then we should find the probability
that x is between 50 and 70 ie. P( 50< x < 70)
For x = 50 , z = (50 - 50) / 15 = 0
For x = 70 , z = (70 - 50) / 15 = 1.33
P( 50< x < 70) = P( 0< z < 1.33) = 0.9082 - 0.5 = 0.4082
The probability that Frank's laptop has a length of time between 50
and 70 hours will be 0.4082.


5. To enter the University you should pass a certain test. The scores
of this test are normally distributed with a mean of 500 and a
standard deviation of 100. Sam would like to be admitted to this

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