Question

Consider the probability that greater than 96 out of 153 DVDs will work correctly. Assume the probability that a given DVD will work correctly is 62%. Approximate the probability using the normal distribution. Round your answer to four decimal places

Answer 1

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

Explanation:

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:

`√(V(X)) = √(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 = (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 = √(V(X))`.

In this problem, we have that:

`n = 153, p = 0.62`

So

`\mu = E(X) = np = 153*0.62 = 94.86`

`\sigma = √(V(X)) = √(np(1-p)) = √(153*0.62*0.38) = 6`

Probability that greater than 96 out of 153 DVDs will work correctly.

97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So

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

`Z = (97 - 94.86)/(6)`

`Z = 0.36`

`Z = 0.36` has a pvalue of 0.6406

1 - 0.6406 = 0.3594

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.