Question

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 3.3% of the general population will live past their 90th birthday. In a graduating class of 796 high school seniors, find the following probabilities. (Round your answers to four decimal places.)

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

  

(b) 30 or more will live beyond their 90th birthday

  

(c) between 25 and 35 will live beyond their 90th birthday

  

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

Answer 1

It is appropriate to use the normal approximation to the binomial because the sample size is very large (796)

p = 0.033

q = 0.967

Mean = np = 796(0.033) = 26.268
Variance = np(1-p) = 796(0.033)(0.967)=25.4011
Standard deviation = 5.0399

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

P( x 15)
Using continuity correction use 14.5 instead of 15.

(14.5-26.268) / 5.0399 = -2.334

P(Z > -2.334) = P(Z < 2.334) = 0.9901

(b) 30 or more will live beyond their 90th birthday

(29.5-26.268) / 5.0399 = 0.6412

P(Z > 0.6412) = =1 − P ( Z < 0.6412 ) = 1 − 0.7389 = 0.2611

(c) between 25 and 35 will live beyond their 90th birthday

P( 25.5 < x < 35.5)

((35.5-26.268) / 5.0399) = 1.8317

((25.5-26.268) / 5.0399) = -0.1523

P( -0.1523 < Z < 1.8317) =   P( Z<1.8317 ) − P (Z<−0.1523 ) = 0.9664 - 0.4404 = 0.526

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

P( x > 40.5)

(40.5-26.268) / 5.0399 = 2.8238

P( Z > 2.8238) = 1 − P ( Z<2.8238 ) = 1−0.9976 = 0.0024