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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.

$X~N(28,5)$

The necessary probability can now be calculated as follows:

So,

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.

$X~N(28,5)$

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.