WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the

Question

2 months ago

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WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the

normal distribution. The mean of the distribution is 15.0 minutes and the standard deviation is 3.5 minutes.
What is the probability that a particular listener will tune in:
a. More than 20 minutes?
b. For 20 minutes or less?
c. Between 10 and 12 minutes?

Mathematics

1 answer:

Inessa [12.5K] · Answer 1 · 2026-03-06T18:14:14+03:00

Answer:

a) $P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

The probability can be determined using the complement rule, with the standard normal distribution, an excel sheet, or a calculator.

$P(z>1.43)=1-P(z$

b) $P(X$

This probability can also be calculated using the normal standard distribution, an excel sheet, or a calculator.

$P(z$

c) $P($

For this one, the probability can likewise be derived from the standard normal distribution, excel, or a calculator, with specific adjustments:

$P(-1.43$

Step-by-step explanation:

Previous concepts

Normal distribution refers to a symmetric probability distribution centered around the mean, indicating that occurrences near the mean are more common than those far from it.

The Z-score is a statistic that represents a value's relationship to the average of a set of values, expressed in terms of how many standard deviations it is away from the mean.

Part a

Let X denote the random variable representing the lengths within a population, and for our case, the distribution for X is as follows:

$X \sim N(15,3.5)$

Where $\mu=15$ and $\sigma=3.5$

We seek the probability:

$P(X>20)$

The most effective way to solve this is by leveraging the normal distribution and the corresponding Z-score:

$z=\frac{x-\mu}{\sigma}$

By applying this formula, we can find the probability:

$P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

Again, this probability can be obtained either using the complement rule, the standard normal distribution, or a calculator.

$P(z>1.43)=1-P(z$

Part b

$P(X$

This probability can also be computed using either the normal standard distribution, an excel sheet, or a calculator.

$P(z$

Part c

$P($

In this case, the probability can similarly be acquired with the help of the standard normal distribution, an excel sheet, or a calculator, with particular adjustments:

$P(-1.43$