Question

Consider the probability that greater than 98 out of 146 houses will not lose power once a year. Assume the probability that a given house will not lose power once a year is 57%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Answer:

0.0054 = 0.54% probability that greater than 98 out of 146 houses will not lose power once a year.

Step-by-step explanation:

I am going to use the normal approximation to the binomial 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:

$E(X) = np$

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:

$n = 146, p = 0.57$

So

$\mu = E(X) = np = 146*0.57 = 83.22$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{146*0.57*0.43} = 5.98$

Probability that greater than 98 out of 146 houses will not lose power once a year.

Using continuity correction, this is $P(X > 98 + 0.5) = P(X > 98.5)$, which is 1 subtracted by the pvalue of Z when X = 98.5. So

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

$Z = \frac{98.5 - 83.22}{5.98}$

$Z = 2.55$

$Z = 2.55$ has a pvalue of 0.9946

1 - 0.9946 = 0.0054

0.0054 = 0.54% probability that greater than 98 out of 146 houses will not lose power once a year.