Where the old folks live Here are a stem plot and numerical summaries of the

percentages of residents aged $65$ and older in the $50$ states and the District of Columbia. Is this distribution of the percent of state residents who are age $65$ and older approximately Normal? Justify your answer based on the graph and the $68–95–99.7$ rule.

The formula used: use of $68-95-99.7$ rule:

According to $68-95-99.7$rule:

In a normal distribution, $68$percent of the data lies within $1$standard deviation of the mean.

In a normal distribution, $95\%$of the data lies within $2$standard deviations of the mean.

A normal distribution has $99.7\%$of its data inside $1$ standard deviation of the mean.

Then

The general Normal density graph is represented as:

Note that

$40$of the $51$observations in the sample are within one standard deviation of the mean.

That means

The sample has $78.43\%$observations within $1$standard deviation.

Between

$\overline{x}-s_x=13.255-1.668=11.587$

And

$\overline{x}+s_x=13.255+1.668=14.923$

Also note that

$48$of the $51$observations in the sample are within$2$ standard deviations of the mean.

That means

The sample has $94.12\%$observations within $2$standard deviations.

Between

$\overline{x}-2s_x=13.255-2(1.668)=9.919$

And

$\overline{x}+2s_x=13.255+2(1.668)=16.591$

Also, note that

The sample has $50$ out of $51$ observations within$3$ standard deviation of the mean.

That means

The sample has $98.04\%$ observations within $3$ standard deviation.

Between

$\overline{x}-3s_x=13.255-3(1.668)=8.251$

And

$\overline{x}+3s_x=13.255+3(1.668)=18.259$

We have that

$78.43\%$is close to $68\%$

$94.12\%$ is close to $95\%$

$98.04\%$ is very close to $99.7\%$

This implies

The data reasonably follows $68-95-99.7$ rule.

Thus,

The distribution is approximately Normal.