Question

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.

On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.

a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30 .

b. Find the ${95}^{\text{th}}$ percentile, and express it in a sentence.

Short answer

(a) The resultant value of $P(x\ge 30)$is $0.3446$.

(b)The percentage of 18 to 34-year-olds who check their Facebook profiles before getting out of bed in the morning is $95\%$ percent.

Step-by-step solution

Part (a) Step 1: Given information

According to the information provided, Facebook's websites present a number of statistics.

It shows that on average $28 percent of people in the 18 to 34 year old age group check their Facebook before getting out of bed, and that this percentage follows a normal distribution with a standard deviation of $5 percent.

Let X represent the variable that 18 to 34 year olds check Facebook before getting out of bed. X is normally distributed with a mean of 28 and a standard deviation of 5.

$X~N(28,5)$

Part (a) Step 2: Explanation

The necessary probability can now be calculated as follows:

$\begin{array}{rcl}P(X& \ge & 30)=1-P(X<30)\\ & =& 1-P\left(\frac{X-\mu }{\sigma }<\frac{30-\mu }{\sigma }\right)\\ & =& 1-P\left(Z<\frac{30-28}{5}\right)\\ & =& 1-P(Z<0.4)\end{array}$

So,

$\begin{array}{rcl}P(X& \ge & 30)=1-0.6554\left\{Froms\tan dardnormaltable\right\}\\ & =& 0.3446\end{array}$

Part (b) Step 1: Given information

According to the information provided, Facebook's websites present a number of statistics.

It shows that on average $28 percent of people in the 18 to 34 year old age group check their Facebook before getting out of bed, and that this percentage follows a normal distribution with a standard deviation of $5 percent.

Let X represent the variable that 18 to 34 year olds check Facebook before getting out of bed. X is normally distributed with a mean of 28 and a standard deviation of 5.

$X~N(28,5)$

Part (b) Step 2: Explanation

The Ti-83 calculator may be used to calculate the ${95}^{th}$percentile.

Use the Ti-83 calculator's invNorm( ) function for this.

To use invNorm(), choose second, then DISTR, then scroll down to the invNorm option and fill in the required information. After that, press the calculator's ENTER key to get the desired result.

The screenshot is as follows:

Therefore, the ${95}^{th}$ percentile is $36.22$ percent.