For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyzed. Find the probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

Step-by-step explanation:

We use the binomial approximation to the normal 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:

The standard deviation of the binomial distribution is:

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:

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:

So

Find the probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

This is 1 subtracted by the pvalue of Z when X = 0.79*92 = 72.68. So

has a pvalue of 0.3121

1 - 0.3121 = 0.6879

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

35 is an outlier

step-by-step explanation: