Question

One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About 14% of American adults are black. The numbers X of blacks in a random sample of 1700 adults should therefore vary with the binomial (n = 1700, p = 0.14) distribution.

(a) What are the mean and standard deviation of X? (Round your standard deviation to three decimal places.)

mean = 238 blacks

SD = 14.307 blacks

(b) Use the Normal approximation to find the probability that the sample will contain between 223 and 253 blacks. (Round your answer to four decimal places.)

P(223 ≤ X ≤ 253) =  

I only need the part to (b) answered. I couldn't figure out this portion of the answer.

Answer 1

Solution :

Given that ,

mean = = 238 blacks

standard deviation = = 14.307 blacks

P(223 x 253)

= P[(223 - 238 / 14.307 ) (x - ) / (253 - 238 / 14.307) ]

= P(-1.05 z 1.05)

= P(z 1.05) - P(z -1.05)

Using z table,  

= 0.8531 - 0.1469

= 0.7062