Question

When conducting an interview with a group of people in order to enter a television program, it is found that 30% of 1500 people do not meet the required requirements. 34 people are interviewed.
a)   What is the probability that less than 24 meet the required requirements?
b)   What is the probability that 14 to 27 do not meet the required requirements?
c)   What is the probability that more than 26 meet the required requirements?
d)   What is the probability that less than 23 or more than 29 do not meet the required requirements?

Answer 1

Sample size, n = 34

P(not meeting the requirement), p = 0.30

q = 1 - p

For random variable X that denotes the number of people who do not meet the requirement,

Normal approximation to binomial: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 34 x 0.30

= 10.2

Standard deviation =

=

= 2.672

a) P(less than 24 meet the requirement) = P(more than 10 do not meet the requirement)

= P(X > 10)

= 1 - P(X < 10.5) (with continuity correction)

= P(Z < (10.5 - 10.2)/2.672)

= P(Z < 0.11)

= 0.5438

b) P(14 to 27 do not meet the requirement) = P(14 X 27)

= P(X < 27.5) - P(X < 13.5)

= P(Z < (27.5 - 10.2)/2.672) - P(Z < (13.5 - 10.2)/2.672)

= P(Z < 6.47) - P(Z < 1.24)

= 1 - 0.8925

= 0.1075

c) P(more than 26 meet the requirement) = P(less than 8 do not meet the requirement)

= P(X < 8)

= P(Z < (8 - 10.2)/2.672)

= P(Z < -0.82)

= 0.2061

d) P(less than 23 or more than 29 do not meet the requirement) = P(X < 23) + P(X > 29)

= P(Z < (22.5 - 10.2)/2.672) + 1 - P(Z < (29 - 10.2)/2.672)

= P(Z < 4.6) + 1 - P(Z < 7.04)

= 1 + 1 - 1

= 1