Question

Use the information in Example 6.10 to answer the following questions.

1. Find the ${30}^{th}$percentile, and interpret it in a complete sentence.
2. What is the probability that the age of a randomly selected smartphone user in the range $13$to $55+$is less than $27$years old.

Short answer

1. The ${30}^{th}$percentile is $29.61$. This percentile value shows that the $30\%$of the total smart phones users of age $13$to ${55}^{+}$years are almost age of $29.61$
2. The probability that the age of a randomly selected smartphone user in the range $13$to $55+$ is less than 27 years old is$0.2382$

Step-by-step solution

Given Information (Part a)

From the information given in the question, we obtain that $X~N(36.9,13.9)$.

Where,

$\mu =36.9$

$\sigma =13.9$

Explanation (Part a)

We can compute the ${30}^{th}$percentile as follow:

$P(x<k)=0.30$

$P\left(\frac{x-\mu }{\sigma }<\frac{k-\mu }{\sigma }\right)=0.30$

$P\left(Z<\frac{k-36.9}{13.9}\right)=0.30$

$\frac{k-36.9}{13.9}=-0.5244\left\{\begin{array}{l}\text{From standard}\\ \text{normal table}\end{array}\right\}$

$k-36.9=-0.5244\times 13.9$

$k=-7.28916+36.9$

$k\approx 29.61$

Given Information (Part b)

The age of a randomly selected smartphone user in the range $13$to $55+$is less than$27$ years old.

Explanation (Part b)

We can compute the required probability as follow:

$P(x<27)=P\left(\frac{x-\mu }{\sigma }<\frac{27-\mu }{\sigma }\right)$

$=P\left(Z<\frac{27-36.9}{13.9}\right)$

$=P(Z<-0.7122)$

$=0.2382\left\{\begin{array}{l}\text{From standard}\\ \text{normal table}\end{array}\right\}$