Question

Consider the probability that more than 91 out of 149 DVDs will work correctly. Assume the probability that a given DVD will work correctly is 66%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Answer:

0.8810 = 88.10% probability that more than 91 out of 149 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:

$n = 149, p = 0.66$

So

$\mu = E(X) = np = 149*0.66 = 98.34$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{149*0.66*0.34} = 5.78$

Consider the probability that more than 91 out of 149 DVDs will work correctly.

Using continuity correction, this is P(X > 91 + 0.5 = 91.5), which is one subtracted by the pvalue of Z when X = 91.5. So

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

$Z = \frac{91.5 - 98.34}{5.78}$

$Z = -1.18$

$Z = -1.18$ has a pvalue of 0.1190

1 - 0.1190 = 0.8810

0.8810 = 88.10% probability that more than 91 out of 149 DVDs will work correctly.