A)
From the information given,
population proportion, p = 0.15
1 - p = 1 - 0.15 = 0.85
sample size, n = 150
We would calculate the standard error by applying the formula,
Standard error = √[p(1 - p)/n]
By substituting the values into the formula, we have
Standard error = √[0.15(0.85)/150]
Standard error = 0.0292
Thus, the sampling distribution of p can be approximated by a normal distribution with mean Ep = 0.15 and standard deviation, σp = 0.0292
B) Given limit = ±0.03, the probability that the sample proportion will be ±0.03 of the population proportion is
p(0.15 - 0.03 < p' < 0.15 + 0.03)
= p(0.12 < p' < 0.18)
Recall,
z = (p' - p)/σp
where
p' is the sample proportion
For p' = 0.12,
z = (0.12 - 0.15)/0.0292
z = - 1.03
For p' = 0.18,
z = (0.18 - 0.15)/0.0292
z = 1.03
0.03/0.0292 = 1.0290
+ 0.03 is 1.0290 standard deviations to the right of p = 0.15
- 0.03 is 1.0290 standard deviations to the left of p = 0.15
We want to find P(0.12 < p' < 0.18)
This is same as
P(- 1.03 < z < 1.03)
From the normal distribution table,
P(z < 1.03) = 0.8485
P(z < -1.03) = 0.1515
P(- 1.03 < z < 1.03) = 0.8485 - 0.1515 = 0.697
the probability that the proportion of the sample is within ±.03 of the proportion of the population is 0.697