Download Quiz Five Solutions

SOLUTIONS
King Fahd University of Petroleum & Minerals
Department of Mathematics & Statistics - STAT-319-Term071-Quiz5
ID:
Sec.:
Serial:
Name:
Q1. The amount of time that a customer spends waiting at an airport check‐in counter is a random variable with mean μ = 8.2 minutes and standard deviation σ = 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. a. What is the sampling distribution of the sample mean? X has approximately normal distribution with 1.
Mean = μ = μ = 8.2 2.
Standard Error = σ X =
X
⎛
Or: X ≈ N ⎜
⎝
μ
= μ = 8.2,σ X
X
σ
1.5
= 0.2143 7
n
⎞
σ
1.5
=
=
= 0.2143 ⎟ 7
n
⎠
=
b. Find the probability that the average time waiting in line for these customers is less than 10 minutes (
P X < 10
)
⎛ X − 8.2 10 − 8.2 ⎞
P⎜
<
⎟ = P ( Z < 8.40 ) ≈ 1 0.2143 ⎠
⎝ 0.2143
Q2. A random sample of size n1 = 16 is selected from a normal population with a mean of μ1 = 75 and a standard deviation of σ 1 = 8 8. A second random sample of size n 2 = 9 is taken from another normal population with mean μ 2 = 70 and standard deviation σ 2 = 12 . Let X 1 and X 2 be the two sample means. Find a. The probability that X 1 − X 2 exceeds 4 2
2
⎛
⎞
(8) (12 )
⎜
= 75 − 70 = 5, σ X 1 − X 2 =
+
= 20 = 4.4721⎟ X 1 −X 2 ∼ N μ
⎜ X 1 −X 2
⎟
16
9
⎝
⎠
⎛ X 1 −X 2 −5 4−5⎞
P X 1 −X 2 > 4 = p⎜
>
⎟ = P ( Z > −0.22 ) = 0.5871 20
20 ⎠
⎝
b. The probability that 3.5 ≤ X 1 − X 2 ≤ 5.5
(
)
⎛ 3.5 − 5 X 1 − X 2 − 5 5.5 − 5 ⎞
≤
≤
P ( 3.5 ≤ X 1 − X 2 ≤ 5.5 ) = P ⎜
⎟
20
20 ⎠
⎝ 20
= P ( −0.34 < Z < 0.11)
= 0.5438 − 0.3669 = 0.1769