Download Name: AMS 7 Homework Quiz 2 Wednesday, October 23

Name:
AMS 7 Homework Quiz 2
Wednesday, October 23, 2013
1. A sample of the blood of 100 people found the following blood types:
Rh Type
O
Positive 39
Negative 6
45
Group
A B AB
35 8
4
5 2
1
40 10
5
86
14
100
(a) (2.5 points) If one person is randomly selected, find the probability of getting
someone who is group B or type Rh-.
P (B or Rh−)
= P (B) + P (Rh−) − P (B and Rh−)
14
2
22
10
+
−
=
= 0.22
=
100 100 100
100
(b) (2.5 points) Given that we select a random person who is Rh-, what is the
probability that they are group B?
P (B|Rh−) =
P (B and Rh−)
2/100
2
=
=
= 0.14
P (Rh−)
14/100
14
(c) (2 points) Suppose that people who are both group B and Rh- visit the blood
clinic at the rate of three per week. Using the Poisson distribution, whamt
would be the mean and standard deviation of the number of people who are
both group B and Rh- who visit the clinic in three weeks?
µ=3·3=9
√
√
σ= µ= 9=3
2. A leatherback turtle has made a nest with 110 eggs. Suppose each egg is independent
and has a probability of 0.85 of hatching.
(a) (2 points) What is the mean and standard deviation of the number of eggs that
hatch?
µ = n · p = 110 · 0.85 = 93.5
p
√
σ = np(1 − p) = 110 · 0.85 · 0.15 = 3.745
(b) (2 points) Would it be unusual for 100 eggs to hatch? Why or why not?
µ + 2σ = 93.5 + 2 · 3.745 = 100.99
So, NO, it is not unusual for 100 eggs to hatch.
3. Assume that adults have IQ scores that are normally distributed with a mean of
100 and a standard deviation of 15.
(a) (3 points) Find the probability that a randomly selected adult gads an IQ that
is less than 110.
X −µ
110 − µ
110 − 100
P (X < 110) = P
<
=P Z<
σ
σ
15
10
= P (Z < ) = P (Z < 0.67) = 0.7486
15
(b) (3 points) Find the probability that a randomly selected adult has an IQ between 95 and 115.
X −µ
115 − µ
95 − µ
<
<
P (95 < X < 115) = P
σ
σ
σ
95 − 100
115 − 100
−5
15
=P
<Z<
= P(
<Z< )
15
15
15
15
= P (Z < 1) − P (Z < −0.33) = 0.8413 − 0.3707 = 0.4706
(c) (3 points) Find P20 , which is the IQ score separating the bottom 20% from the
top 80%.
We want to find
P (X < P20 ) = 0.20
We know that
P (Z < −0.845) = 0.20
⇒ P (Zσ + µ < −0.845σ + µ) = 0.20 (∗∗)
⇒ P (X < −0.845 · 15 + 100) = 0.20
⇒ P (X < 87.325) = 0.20
So P20 = 87.325
(∗∗) Z =
X −µ
⇒ X = Zσ + µ
σ