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 µ = 339 days, and σ = 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?
Over
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?
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 µ = 339 days, and σ = 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.
4
U
20
8
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
14
U
20
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
U
15 20
= 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)
=
4
6
8
20
1
(8 – 6)
16
= 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