Question

An employer has determined that the proportion of all its employees who drink coffee is 85%.

a. What is the probability that in a random sample of 200 employees, less than 80% drink coffee?

b. What is the probability that in a random sample of 200 employees, more than 90% drink coffee?

c. Interpret the probability you have calculated for part-b above using your own words.

Answer 1

We know this is binomial distribution with p=0.85 and n=200

p is the proportion of people who drinks coffee

a. We have to find out P(x<160)

We will use normal approximation

X~N(E(X),Var(X))

E(X) = np = 200*0.85 = 170

Var(X) = np(1-p) = 200*0.85*0.15 = 25.5

X ~ N(170,25.5)

When we convert discrete distribution to continuous distribution we have to do continuity correction, So, now we have to find P(X<159.5) We will convert it to standard normal

P(Z<(159.5-170)/25.5^0.5)

P(Z<-2.08) = P(Z>2.08)

We can find its value from Standard distribution table or Z table

P(Z>2.08) = 1- P(Z<2.08) (As we are given less than value in Z table)

= 1 - 0.98124

= 0.01876 is the answer

b. Now we have to find P(X>180)

With continuity correction

P(X>180.5)

Converting it to standard normal

P(Z>(180.5-170)/25.5^0.5)

P(Z>2.08) = 1 - (PZ<2.08)

= 1 - 0.98124

= 0.01876 is the answer

c. We can say from part b that there are 0.01876 chances that more than 180 people drinks coffee