Question

It is commonly known that approximately 10% of the population is left-handed. Nike Golf would like to estimate the proportion of left-handed golfers in order to plan the schedule of golf club production. Assume Nike collected a sample of 200 golfers.

1. What is the probability that more than 12% of the golfers in the sample are left-handed?

2. What is the probability that less than 14% of the golfers in the sample are left-handed?
3. What is the probability that between 6% and 9% of the golfers in the sample are left-handed?
4. Suppose the sample that Nike collected had nine left-handed golfers. What conclusions should Nike draw about the population of left-handed golfers

4. Suppose the sample that Nike collected had nine left-handed golfers. What conclusions should Nike draw about the population of left-handed golfers?

Answer 1

Answer)

N = 200

P = 0.1

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 20

N*(1-p) = 180

Both the conditions are met so we can use standard normal z table to estimate the probability.

z = (observed p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

A)

Observed P = 0.12

Claimed P = 0.1

N = 200.

After substitution,

Z = 0.94

From z table, P(z>0.94) = 0.1736

B)

P(p<0.14) = P(z<1.89) = 0.9706

C)

P(0.06 < p < 0.09) = p(p<0.09) - p(p<0.06)

= p(z<-0.47) - p(z<-1.89)

= 0.3192 - 0.0294

= 0.2898

D)

Sample proportion is 0.045 (9/200)

Claimed P = 0.1

N = 200

After substitution,

Z = -2.59

From z table, P(z<-2.59) = 0.0048

As the probability is extremely small less than 0.05.

We can conclude that population proportion is not 10% (<10%).