Question

According to government data, the probability than an adult never had the flu is 19%. You randomly select 70 adults and ask if he or she ever had the flu. Decide whether you can use the normal distribution to approximate the binomial distribution, If so, find the mean and standard deviation, If not, explain why. Round to the nearest hundredth when necessary.

Answer 1

Answer:

Since $np \geq 10$ and $n(1-p) \geq 10$, the normal distribution can be used to approximate the binomial distribution.

The mean is 13.3 and the standard deviation is 3.28.

Step-by-step explanation:

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$, if $np \geq 10$ and $n(1-p) \geq 10$.

The probability than an adult never had the flu is 19%.

This means that $p = 0.19$

You randomly select 70 adults and ask if he or she ever had the flu.

This means that $n = 70$

Decide whether you can use the normal distribution to approximate the binomial distribution

$np = 70*0.19 = 13.3 \geq 10$

$n(1-p) = 70*0.81 = 56.7 \geq 10$

Since $np \geq 10$ and $n(1-p) \geq 10$, the normal distribution can be used to approximate the binomial distribution.

Mean:

$\mu = E(X) = np = 70*0.19 = 13.3$

Standard deviation:

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{70*0.19*0.81} = 3.28$

The mean is 13.3 and the standard deviation is 3.28.