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

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