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

Answer:

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

Step-by-step 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:

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

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:

$n = 153, p = 0.62$

So

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

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.62*0.38} = 6$

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

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

$Z = \frac{97 - 94.86}{6}$

$Z = 0.36$

$Z = 0.36$ 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.