DS 520-Week 4 1
DS 520-Week 4
Student’s name
Institutional affiliation
Course
Dr. Professor’s name
Due Date
, DS 520-Week 4 2
5.76 The cost of Internet access
We know the population mean µ = 68 and the standard deviation σ = 22.
The sampling distribution of the sample mean, for the sample size (n) 500, is
μ ( x ¿ = 𝜇 = 68 and the standard deviation σ ( x ¿ = σ/√n
σ ( x ¿ = 22/√500 = 22/22.306 = 0.986
Since n > 30, the Central Limit Theorem applies, thus we can assume the sample mean is
normally distributed with 𝑁(68, 0.986).
For 𝑥̅ = 70, 𝑧 = 70−68
= 2/0.986 = 2.02
According to the standard normal table,
𝑃(𝑋 > 70) = 𝑃(𝑍 > 2.02) = 1 − 𝑃(𝑍 < 2.02) = 1 − 0.97831 = 0.02169
5.82 Genetics of peas
To solve this problem the binomial law is applied, where P(X= x) = nCx* (p)x * (q)(n-x)
(a) p= 3/4= 0.75,
q= 1-0.75 = 0.25
Probability that exactly nine out of twelve plants have red blossoms
x = 9, n = 12 and nCx = n!/[x! (n-x)!]
P(X=9) = 12C9* (0.75)9 * (0.25)3 = 220 * 0.075 * 0.0156
= 0.2574
DS 520-Week 4
Student’s name
Institutional affiliation
Course
Dr. Professor’s name
Due Date
, DS 520-Week 4 2
5.76 The cost of Internet access
We know the population mean µ = 68 and the standard deviation σ = 22.
The sampling distribution of the sample mean, for the sample size (n) 500, is
μ ( x ¿ = 𝜇 = 68 and the standard deviation σ ( x ¿ = σ/√n
σ ( x ¿ = 22/√500 = 22/22.306 = 0.986
Since n > 30, the Central Limit Theorem applies, thus we can assume the sample mean is
normally distributed with 𝑁(68, 0.986).
For 𝑥̅ = 70, 𝑧 = 70−68
= 2/0.986 = 2.02
According to the standard normal table,
𝑃(𝑋 > 70) = 𝑃(𝑍 > 2.02) = 1 − 𝑃(𝑍 < 2.02) = 1 − 0.97831 = 0.02169
5.82 Genetics of peas
To solve this problem the binomial law is applied, where P(X= x) = nCx* (p)x * (q)(n-x)
(a) p= 3/4= 0.75,
q= 1-0.75 = 0.25
Probability that exactly nine out of twelve plants have red blossoms
x = 9, n = 12 and nCx = n!/[x! (n-x)!]
P(X=9) = 12C9* (0.75)9 * (0.25)3 = 220 * 0.075 * 0.0156
= 0.2574
