Question

When telephone subscribers call from the National Magazine Subscription Company, 18% of the people who answer stay on the line for more than one minute. If 800 people are called in a day, find the probability that a. at least 150 stay on the line for more than one minute. (Use normal approximation to binomial). b. more than 200 stay on the line. (Use Normal approximation to Binomial).

Answer 1

Answer:

a) 30.50% probability that  at least 150 stay on the line for more than one minute.

b) 0% probability that more than 200 stay on the line for more than one minute.

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:

$n = 800, p = 0.18$

So

$\mu = E(X) = np = 800*0.18 = 144$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{800*0.18*0.82} = 10.87$

a. at least 150 stay on the line for more than one minute.

Using continuity correction, $P(X \geq 150 - 0.5) = P(X \geq 149.5)$, which is 1 subtracted by the pvalue of Z when X = 149.5. So

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

$Z = \frac{149.4 - 144}{10.87}$

$Z = 0.51$

$Z = 0.51$ has a pvalue of 0.6950

1 - 0.6950 = 0.3050

30.50% probability that  at least 150 stay on the line for more than one minute.

b. more than 200 stay on the line.

Using continuity correction, $P(X \geq 200 + 0.5) = P(X \geq 200.5)$, which is 1 subtracted by the pvalue of Z when X = 200.5. So

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

$Z = \frac{200.5 - 144}{10.87}$

$Z = 5.2$

$Z = 5.2$ has a pvalue of 1

1 - 1 = 0

0% probability that more than 200 stay on the line for more than one minute.