Question

Consider the probability that at most 85 out of 136 DVDs will work correctly. Assume the probability that a given DVD will work correctly is 52%. Specify whether the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

Answer 1

Answer:

Since both $np \geq 10$ and $n(1-p) \geq 10$, the necessary conditions are satisfied.

0.9945 = 99.45% probability that at most 85 out of 136 DVDs will work correctly.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

Binomial probability distribution

Probability of exactly x successes 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 distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

Assume the probability that a given DVD will work correctly is 52%.

This means that $p = 0.52$

136 DVDs

This means that $n = 136$

Test the conditions:

$np = 136*0.52 = 70.72 \geq 10$

$n(1-p) = 136*0.48 = 65.28 \geq 10$

Since both $np \geq 10$ and $n(1-p) \geq 10$, the necessary conditions are satisfied.

Mean and standard deviation:

$\mu = E(X) = np = 136*0.52 = 70.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{136*0.52*0.48} = 5.83$

Consider the probability that at most 85 out of 136 DVDs will work correctly.

Using continuity correction, this is $P(X \leq 85 + 0.5) = P(X \leq 85.5)$, which is the p-value of Z when X = 85.5. So

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

$Z = \frac{85.5 - 70.72}{5.83}$

$Z = 2.54$

$Z = 2.54$ has a p-value of 0.9945.

0.9945 = 99.45% probability that at most 85 out of 136 DVDs will work correctly.