In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

It is estimated that 3.8% of the general population will live past their 90th birthday. In a graduating class of 723 high school seniors, find the following probabilities. (Round your answers to four decimal places.)

Question Continuation

Find the following probabilities. (Round your answers to four decimal places.)

(a) 15 or more will live beyond their 90th birthday

(b) more than 40 will live beyond their 90th birthday

a. 0.9948

b. 0.0068

Step-by-step explanation:

Given

Number of graduating students, n = 723

Probability, p = 3.8% = 0.038

q = 1 - p = 1 - 0.038 = 0.962

Calculating the mean

Mean, u = np

u = 723 * 0.038

u = 27.474

Calculating the standard deviation

σ = √npq

σ = √(723 * 0.038 * 0.962)

σ = 5.14101040652516088641079

σ = 5.141 --- Approximated

We'll solve the following using z = (x-u)/σ

a.

x = 14.5 (I.e Atleast 15)

u = 27.474

σ = 5.141

Z = (14.5 - 27.474)/5.141

Z = −2.5236335343318420540750

Z = -2.52

From the binomial distribution table

The corresponding value at z = -2.52 is 0.0052

So, the probability that 15 or more will live beyond their 90th birthday = 1 - 0.0052

= 0.9948

b.

For more than 40

x = 40.5

u = 27.474

σ = 5.141

Z = (40.5 - 27.474)/5.141

Z = 2.53374829799649873565454

Z = 2.53

From the binomial distribution table

The corresponding value at z = 2.53 is 0.9932

So, the probability that more than 40 will live beyond their 90th birthday = 1 - 0.9932

= 0.0068

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