Answer:
1.04% probability that a simple random sample of 100 of these customers would contain at least 57 persons who preferred small curd cottage cheese
Step-by-step explanation:
I am going to use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that
,
.
In this problem, we have that:
![p = 0.45, n = 100](https://tex.z-dn.net/?f=p%20%3D%200.45%2C%20n%20%3D%20100)
![\mu = E(X) = 100*0.45 = 45](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20100%2A0.45%20%3D%2045)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.45*0.55} = 4.9749](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B100%2A0.45%2A0.55%7D%20%3D%204.9749)
What is the approximate probability that a simple random sample of 100 of these customers would contain at least 57 persons who preferred small curd cottage cheese?
Using continuity correction, this is
, which is 1 subtracted by the pvalue of Z when X = 56.5. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{56.5 - 45}{4.9749}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B56.5%20-%2045%7D%7B4.9749%7D)
![Z = 2.31](https://tex.z-dn.net/?f=Z%20%3D%202.31)
has a pvalue of 0.9896
1 - 0.9896 = 0.0104
1.04% probability that a simple random sample of 100 of these customers would contain at least 57 persons who preferred small curd cottage cheese