Question

A research institute reports that 65% of workers reported that they and/or their spouse had saved some moneh for retirement. Complete the parts below.

a. if a random sample of 50 workers is taken, what is the probability that fewer than 29 workers and/or their spouses have saved some money for retirement?

b. If a random sample of 60 workers is taken, what is the probability that more than 48 workers and/or their spouses have saved money for retirement

Answer 1

Solution:

Given that,

P = 65% = 0.65

1 - P = 1 - 0.65 = 0.35

a)

n = 50

Here, BIN ( n , P ) that is , BIN (50 , 0.65)

then,

n*p = 50 * 0.65 = 32.5 > 5

n(1- P) = 50 * 0.35 = 17.5 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 32.5

Standard deviation = =n*p*(1-p)= 50*0.65*0.35 = 11.375

We use countinuity correction factor

P(X < a ) = P(X < a - 0.5)

P(x < 28.5) = P((x - ) / < (28.5 - 32.5) / 11.375)

= P(z < -1.186)

Probability = 0.1178

b)

n = 60

Here, BIN ( n , P ) that is , BIN (60 , 0.65)

then,

n*p = 60*0.65 = 39 > 5

n(1- P) = 60 * 0.35 = 21 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 39

Standard deviation = =60*0.65*0.35= 13.65

We using countinuity correction factor

P(x > a ) = P( X > a + 0.5)

P(x > 48.5) = 1 - P(x < 48.5)

= 1 - P((x - ) / < (48.5 - 39) / 13.65)

= 1 - P(z < 2.571)

= 1 - 0.9949   

= 0.0051

Probability = 0.0051