Question

It is estimated that 4.5% of the general population will live past their 90th birthday. Suppose a random sample of 753 high school seniors is selected.

a. Describe the sampling distribution of ?̂.

b. What is the probability that from this random sample taken, the sample proportion of these high school seniors who will live past their 90th birthday is between 0.043 and 0.056?

Answer 1

Given,

Proportion of people will live past their 90th Birthday : p = 4.5/100 =0.045

Sample size : Number of high school seniors selected in the random sample : n=753

a.

Sampling distribution of follows normal distribution with mean: = 0.045 and standard deviation :

Sampling distribution of follows normal distribution with mean =0.045 ; standard deviation :0.0181

b.

probability that from this random sample taken, the sample proportion of these high school seniors who will live past their 90th birthday is between 0.043 and 0.056 = P(0.043 < < 0.056)

P(0.043 < < 0.056) = P( < 0.056) - P( < 0.043)

Z-score for 0.056 = (0.056-0.045)/0.0181 = 0.61

Z-score for 0.043 = (0.043-0.045)/0.0181 = -0.11

From standard normal tables,

P(Z<0.61) =0.7291 ;  P(Z<-0.11) = 0.4562

P( < 0.056) = P(Z<0.61) =0.7291

P( < 0.043) = P(Z<-0.11) = 0.4562

P(0.043 < < 0.056) = P( < 0.056) - P( < 0.043) = 0.7291-0.4562=0.2729

Probability that from this random sample taken, the sample proportion of these high school seniors who will live past their 90th birthday is between 0.043 and 0.056 = 0.2729