DS 520 Week 4

DS 520-Week 4 1




DS 520-Week 4



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, 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