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?

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

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option