The distribution of grade point averages for a certain college is approximately Normal with a mean of $2.5$and a standard deviation of $0.6$. According to the $68–95–99.7$rule, within which interval would we expect to find approximately $81.5\%$of all GPAs for students at this college?

(a)$(0.7,3.1)$

(b)$(1.3,3.7)$

(c)role="math" localid="1650374007246" $(1.9,3.7)$

(d)$(1.9,4.2)$

(e)$(0.7,4.2)$

Given

$\mu =Mean=2.5$

$\sigma =Standarddeviation=0.6$

The $z$-score is the difference between the mean and the standard deviation.

$z=\frac{x-\mu }{\sigma }=\frac{0.7-2.5}{0.6}\approx -3.00$

$z=\frac{x-\mu }{\sigma }=\frac{1.3-2.5}{0.6}\approx -2.00$

$z=\frac{x-\mu }{\sigma }=\frac{1.9-2.5}{0.6}\approx -1.00$

$z=\frac{x-\mu }{\sigma }=\frac{3.1-2.5}{0.6}\approx 1.00$

$z=\frac{x-\mu }{\sigma }=\frac{3.7-2.5}{0.6}\approx 2.00$

$z=\frac{x-\mu }{\sigma }=\frac{4.3-2.5}{0.6}\approx 3.00$

Using the normal probability table in the appendix, calculate the corresponding probability. $P(Z<-3.00)$is given in the row starting with $-3.0$and in the column starting with $.00$of the normal probability table (similar for the other $z$-scores).

localid="1650381870931" $$\begin{gathered} P(0.7<X<3.1)=P(-3.00<Z<1.00) \\ =P(Z<1.00)-P(Z<-3.00) \\ =0.8413-0.0013 \\ =0.8400 \\ =84.00\mathrm{\%} \end{gathered}$$

localid="1650381875270" $$\begin{gathered} P(1.3<X<3.7)=P(-2.00<Z<2.00) \\ =P(Z<2.00)-P(Z<-2.00) \\ =0.9772-0.0228 \\ =0.9544 \\ =95.44\mathrm{\%} \end{gathered}$$

localid="1650381879885" $$\begin{gathered} P(1.9<X<3.7)=P(-1.00<Z<2.00) \\ =P(Z<2.00)-P(Z<-1.00) \\ =0.9772-0.1587 \\ =0.8185 \\ =81.85\mathrm{\%} \end{gathered}$$

localid="1650381885683" $$\begin{gathered} P(1.9<X<4.3)=P(-1.00<Z<3.00) \\ =P(Z<3.00)-P(Z<-1.00) \\ =0.9987-0.1587 \\ =0.8400 \\ =84.00\mathrm{\%} \end{gathered}$$

localid="1650381891903" $$\begin{gathered} P(0.7<X<4.3)=P(-3.00<Z<3.00) \\ =P(Z<3.00)-P(Z<-3.00) \\ =0.9987-0.0013 \\ =0.9974 \\ =99.74\mathrm{\%} \end{gathered}$$

We then note that the interval $(1.9,3.7)$has a probability of $81.85\%$, which is approximately $81.5\%$.