Based on labor​ statistics, 21% of workers in a particular profession are male. Complete parts a through e below based on a random sample of 8 workers in this profession. (round answers to four decimal places)

a. What is the probability that exactly one worker in the sample is male​?

b. What is the probability that fewer than 4 workers in the sample are male​?

c. What is the probability that more than 2 workers in the sample are male​?

d. What are the mean and standard deviation for this​ distribution?

Respuesta :

Answer:

a) 0.3226

b) 0.9340

c) 0.5257

d) mean=1.68 workers , standard deviation=1.15 workers

Step-by-step explanation:

since each worker's gender is independent from the others , then defining the random variable X= getting x male workers out of the sample of 8 workers , we know that P(X) has a binomial distribution , where

P(X)=n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= sample size = 8

p= probability that a worker is male = 0.21

x= x workers are male

then

a) P(X=1) = 8!/((8-1)!*1!)*(0.21)^1*(1-0.21)^(8-1) = 0.3226

b)  P(X<4) =  P(X=0) + P(X=1)+ P(X=2)+ P(X=3) + P(X=4)

in order to avoid doing the calculus for each term we can use the cumulative probability distribution , whose results can be found in tables. Then

P(X<4)= F(4) = 0.9340

c) P(X>2) = 1- P(X≤1) = 1- F(1) = 1- 0.4743 = 0.5257

d) the mean for a binomial distribution is

E(X)= n*p =  8*0.21 = 1.68 workers

and the standard deviation is

σ(X)= √[n*p*(1-p)]= √[8*0.21*0.79]= 1.15 workers