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A bottle maker believes that 33% of his bottles are defective. If the bottle maker is right, what is the probability that the proportion of defective bottles in a sample of 730 bottles would differ from the population proportion by greater than 3%

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Answer:

8.54% probability that the proportion of defective bottles in a sample of 730 bottles would differ from the population proportion by greater than 3%

Explanation:

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 = (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.

For a proportion p in a sample of size n, we have that
\mu = p, \sigma = \sqrt{(p(1-p))/(n)}

In this problem


\mu = 0.33, \sigma = \sqrt{(0.33*0.67)/(730)} = 0.0174

What is the probability that the proportion of defective bottles in a sample of 730 bottles would differ from the population proportion by greater than 3%

Either less than 0.33 - 0.03 = 0.3 or greater than 0.33 + 0.03 = 0.36.

Since the normal distribution is symmetric, and both these values are the same distance from the mean, we can find any of these probabilities and multiply by 2, since they are equal.

Probability of being lower than 0.3.

pvalue of Z when X = 0.3. So


Z = (X - \mu)/(\sigma)


Z = (0.3 - 0.33)/(0.0174)


Z = -1.72


Z = -1.72 has a pvalue of 0.0427

2*0.0427 = 0.0854

8.54% probability that the proportion of defective bottles in a sample of 730 bottles would differ from the population proportion by greater than 3%

User Carolin
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