Question

Although studies continue to show smoking leads to significant health problems, 20% of adults in the United States smoke. Consider a group of 300 adults.

If required, round your answers to four decimal places. Use Table 1 in Appendix B. If your answer is zero, enter "0".

c. What is the probability that from 35 to 70 smoke? Used continuity correction factor.

A binomial probability distribution has p = .20 and n = 100.

If required, round your answers to four decimal places. Use “Continuity correction factor, if necessary”. Use Table 1 in Appendix B.

c. What is the probability of exactly 22 successes?

Answer 1

Answer:

Given,

sample n = 300

p = 0.20

Mean = np

= 300*0.20

= 60

Standard deviation = sqrt(npq)

= sqrt(300*0.20*0.80)

= 6.93

a)

P(35 < X < 70) = P(34.5 < X < 70.5)

= P((34.5 - 60)/6.93 < (x-u)/s < (70.5 - 60)/6.93)

= P(-3.68 < z < 1.52)

= P(z < 1.52) - P(z < - 3.68)

= 0.9357445 - 0.0001166 [since from z table]

= 0.9356

b)

n = 100

p = 0.20

Binomial distribution P(X = r) = nCr*p^r*q^(n-r)

nCr = n!/(n-r)!*r!

P(X = 22) = 100C22 * 0.20^22 * 0.80^(100-22)

= 7332066885177656269200*0.20^22*0.80^78

= 0.0849