Question

in the following problem check that it is appropriate to use a normal approximation to the binomial. then use the normal distribution to estimate the request of probabilities. it is estimated at 3.5% of the general population live past her 90th birthday. Space in a graduating year of 751 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 it be on their 90th birthday

d) More than 40 will it be on their 90th birthday

Answer 1

n = number of high school seniors selected = 751

p =P ( a senior will live beyond their 90^th birthday) = 0.035

Let the random variable X is defined as

X : Number of seniors will will live beyond their 90^th birthday in a sample

X ~ Bin ( n = 751, p = 0.035)

E(X) = n*p = 751 *0.035 = 26.285

SD(X) = sqrt(n*p*(1-p)) = sqrt(751 * 0.035 * (1-0.035)) = 5.0364

Since n is large and p is very small.

By using normal approximation to binomial distribution

a) P ( 15 or more will live beyond their 90 ^th birthday) = P ( X >= 15)

=

= P ( Z >= -2.2407)

From normal probability table

P ( Z > = -2.2407) = 0.9875

P ( 15 or more will live beyond their 90 th birthday) = 0.9875

b) P ( 30 or more will live beyond their 90 ^th birthday) =P ( X> =30)

= P( Z>= 0.7376)

From normal probability table

P ( Z>= 0.7376) = 0.2304

P ( 30 or more will live beyond their 90 ^th birthday) = 0.2304

c) P ( Between 25 and 35 will it be on their 90^th birthday) = P ( 25 < X < 35)

= P ( -0.2551 < Z < 1.7304)

=P ( Z< 1.7304) - P ( Z < -0.2551)

From normal probability tale

P ( Z < 1.7304) = 0.9582 and P ( Z < -0.2551) = 0.3993

P ( Between 25 and 35 will it be on their 90^th birthday) = 0.9582 - 0.3993 = 0.5589

d) P ( more than 40 will t be on their 90^th birthday) = P ( X > 40)

= P ( Z > 2.7231)

From normal probability table

P ( Z > 2.7231)=0.0032

P ( more than 40 will t be on their 90^th birthday) = 0.0032.