Question

Forty-five percent of all customers of a dairy company prefer large curd cottage cheese to small curd cottage cheese. 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?

Answer 1

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$p = 0.45, n = 100$

$\mu = E(X) = 100*0.45 = 45$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.45*0.55} = 4.9749$

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 $P(X \geq 57 - 0.5) = P(X \geq 56.5)$, which is 1 subtracted by the pvalue of Z when X = 56.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{56.5 - 45}{4.9749}$

$Z = 2.31$

$Z = 2.31$ 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