Normal and Uniform Distribution Problems Associated with Section 4.2
Answers on pages 3 and 4.
1. From section 1.3 we had the problem describing the length of horse pregnancies. The length of horse
pregnancies is approximately normal, with  = 339 days, and  = 3 days. Let the random variable H denote the
length of a horse pregnancy is days. For the problems given below, draw a graph of the situation, and use
function notation to communicate what you are doing.
a. A horse is found to be pregnant, what is the probability of the length of the pregnancy lasting less than 336
days or more than 345 days?
b. A horse is found to be pregnant. The owner of the horse wants the pregnancy to last less than 240 days. The
doctor would like the pregnancy to last longer than 335 days. What is the probability of satisfying both events?
c. A horse is found to be pregnant. The owner of the horse wants the pregnancy to last less than 340 days. The
doctor would like the pregnancy to last longer than 335 days. What is the probability of satisfying both events?
2. The random number generator in Excel is set to produce numbers at random from 4 to 20 using a uniform
distribution. Let the random variable U denote the value of a uniform random number. For the problems given
below, draw a graph of the situation, and use function notation to communicate what you are doing.
Over
a. What is the probability of getting a value less than 8?
b. Calculate P(U > 6 AND U < 14)
c. Calculate P(U > 15 or U < 8)?
d. Define event A to be a number is less than 8 (probability calculated in “a”). Define event B to be a number is
in the interval 6 < U < 14 (probability calculated in “b”). Are the events A and B independent? Use the formula
P(A and B) = P(A)P(B) to make the determination. Note that equality is only true when we have independence.
Answers to the Problems
Normal and Uniform Distribution Problems Associated with Section 4.2
1. From section 1.3 we had the problem describing the length of horse pregnancies. The length of horse
pregnancies is approximately normal, with  = 339 days, and  = 3 days. Let the random variable H denote the
length of a horse pregnancy is days. For the problems given below, draw a graph of the situation, and use
function notation to communicate what you are doing.
a. A horse is found to be pregnant, what is the probability of the length of the pregnancy lasting less than 336
days or more than 345 days?
P(H < 336 OR H > 345) = P(Z < -1 OR Z > 2)
= P(Z < -1) + P(Z > 2)
= 0.16 + 0.025 using 68-95-99.7 rule
= 0.185
= 0.1587 +0.0228 using tables or software
= 0.1815
b. A horse is found to be pregnant. The owner of the horse wants the pregnancy to last less than 240 days. The
doctor would like the pregnancy to last longer than 335 days. What is the probability of satisfying both events?
P(H < 240 AND H > 335) = 0 There is no way to satisfy
both events simultaneously.
c. A horse is found to be pregnant. The owner of the horse wants the pregnancy to last less than 340 days. The
doctor would like the pregnancy to last longer than 335 days. What is the probability of satisfying both events?
P(H < 340 AND H > 335) = P(335 < H < 340)


= PZ <
340 - 339  
335 - 339 
  PZ <

3  
3 
= P(Z < 0.3333) – P(Z < -1.3333)
= 0.6306 0.0912
= 0.5394
2. The random number generator in Excel is set to produce numbers at random from 4 to 20 using a uniform
distribution. Let the random variable U denote the value of a uniform random number. For the problems given
below, draw a graph of the situation, and use function notation to communicate what you are doing.
a. What is the probability of getting a value less than 8?
P(4 < U < 8) =
1
(8 – 4)
16
= 0.25.
b. Calculate P(U > 6 AND U < 14)
P(U > 6 AND U < 14) = P(6 < U < 14)
=
1
(14 – 6)
16
= 0.5.
4
6
20
14
U
c. Calculate P(U > 15 or U < 8)?
P(U > 15 or U < 8) = P(U < 8) + P(U > 15)
= 0.25 +
1
(20 – 15)
16
= 0.25 + 0.3125
4
8
15
20
U
= 0.5625
d. Define event A to be a number is less than 8 (probability calculated in “a”). Define event B to be a number is
in the interval 6 < U < 14 (probability calculated in “b”). Are the events A and B independent? Use the formula
P(A and B) = P(A)P(B) to make the determination. Note that equality is only true when we have independence.
P(U < 8) = 0.25 P(6 < U < 14) = 0.5
P(U < 8)P(6 < U < 14) = 0.5(0.25)
= 0.125
P(U < 8 AND 6 < U < 14) = P(6 < U < 8)
1
=
(8 – 6)
16
4 6 8
20
= 0.125
Since P(U < 8)P(6 < U < 14) = P(U < 8 AND 6 < U < 14) we have independence. We could have used the
formula P(A |B) = P(A) to show we have independence.
P(U < 8 | 6 < U < 14) =
0.125
= 0.25 = P(U < 8).
0.5
U