Question

1a. What is the area under standard normal density curve to the left of 1.96?

1b. What is the 99th percentile of standard normal density curve?

1c. If 5-year survival rate is 24% for a certain disease, what is the probability that at least 3 out of 10 patients will survive more than 5 years? how about at least 30 out of 100 patients? (use central limit theorem)

1d. What is the name of the theorem that tells us the distribution of average of samples when sample size is large?

Answer 1

1a. What is the area under standard normal density curve to the left of 1.96?

Here, we have to find P(Z<1.96) for a standard normal density.

So, by using z-table, we have

P(Z<1.96) = 0.975002

Required answer: 0.975002

1b. What is the 99th percentile of standard normal density curve?

Here, we have to find the value of z for a standard normal density such that P(Z<z) = 0.99

So, by using z-table, we have

Z = 2.326348

99^th percentile = 2.326348

1c. If 5-year survival rate is 24% for a certain disease, what is the probability that at least 3 out of 10 patients will survive more than 5 years? how about at least 30 out of 100 patients? (use central limit theorem)

First we have to find P(X≥3) by using binominal distribution with n = 10, p = 0.24

P(X≥3) = 1 – P(X≤2)

P(X≤2) = P(X=0) + P(X=1) + P(X=2)

P(X=x) = nCx*p^x*q^(n – x)

Where, q = 1 – p = 1 – 0.24 = 0.76

P(X=0) = 10C0*0.24^0*0.76^(10 – 0)

P(X=0) = 1*0.24^0*0.76^10

P(X=0) = 0.064289

P(X=1) = 10C1*0.24^1*0.76^(10 – 1)

P(X=1) = 10*0.24^1*0.76^9

P(X=1) = 0.203018

P(X=2) = 10C2*0.24^2*0.76^(10 – 2)

P(X=2) = 45*0.24^2*0.76^8

P(X=2) = 0.288499

P(X≤2) = P(X=0) + P(X=1) + P(X=2)

P(X≤2) = 0.064289 + 0.203018 + 0.288499

P(X≤2) = 0.555806

P(X≥3) = 1 – P(X≤2)

P(X≥3) = 1 – 0.555806

P(X≥3) = 0.444194

Required probability = 0.444194

Now, we have to find P(X≥30) by using normal approximation with n = 100, p = 0.24, q = 0.76

Mean = np = 100*.24 = 24

SD = sqrt(npq) = sqrt(100*0.24*0.76) = 4.270831

P(X≥30) = P(X>29.5) (by using continuity correction)

P(X>29.5) = 1 – P(X<29.5)

Z = (X – mean) / SD

Z = (29.5 - 24)/ 4.270831

Z = 1.287806

P(Z<1.287806) = P(X<29.5) = 0.901093

(by using z-table)

P(X>29.5) = 1 – P(X<29.5)

P(X>29.5) = 1 – 0.901093

P(X>29.5) = 0.098907

Required probability = 0.098907

1d. What is the name of the theorem that tells us the distribution of average of samples when sample size is large?

The name of the required theorem is the central limit theorem or normal approximation which states that the sampling distribution of the mean follows an approximate normal distribution.