Question

Suppose 75% of all farms in a certain state produce corn. A simple random sample of 60 the state's farms is drawn. (Use technology. Round your answers to three decimal places.) (a) Find the probability that 50 or fewer farms in this sample produce corn. (b) Find the probability that the number of farms producing corn is between 40 and 50, inclusive. That is, find P(40 ≤ x ≤ 50).

Answer 1

75% of all farms in a certain state produce corn. So, P= 0.75

A simple random sample of 60 state's farms is drawn. So sample size n=60

We can use binomial probability distribution for this scenario. Where x= random variable represents the number of farms in the sample that produce corn.

(a) the probability that 50 or fewer farms in this sample produce corn is p(x<=50)

enter the value of p in first cell and n value in second cell and x value in 3rd cell.

The highlighted is the required probability P(x<=50)=0.9548(rounded to 4 decimal places).

(b) The probability that the number of farms producing corn is between 40 and 50 is P(40<=x<=50)=p(x<=50)-p(x<=40)

we already found that P(x<=50) = 0.9548

Now find P(x<=40) using the same

As explained above P(x<= 40) found to be 0.0925 (rounded to 4 decimal places).

Now P(40<=x<=50)= 0.9548-0.0925=0.8623.

Therefore P(40<=x<=50) = 0.8623.