Solve the Problem $65.2$ Inches and That the Standard Deviation Is $2.8$

Solve the problem.
-A northeastern college has an enrollment of 2835 female students. Records show that the mean height of these students is $65.2$ inches and that the standard deviation is $2.8$ inches. The table shows frequency and relative-frequency data for these heights. If you assume that the distribution of heights is approximately normal, then you can use the table to estimate areas under the associated normal curve (that is, under the normal curve that has parameters $\mu = 65.2$ and $\sigma = 2.8$ ) . Making this assumption, estimate the area under the associated normal curve between 65 and 70 . $\begin{array}{c|c|c}\begin{array}{c}\text { Height } \\\text { (inches) }\end{array} & \text { Freq. } & \begin{array}{c}\text { Relative } \\\text { freq. }\end{array} \\\hline \text { 57-under 58 } & 16 & 0.0056 \\58 \text {-under 59 } & 25 & 0.0088 \\59 \text {-under 60 } & 56 & 0.0198 \\\text { 60-under 61 } & 96 & 0.0339 \\\text { 61-under 62 } & 178 & 0.0628 \\\text { 62-under 63 } & 263 & 0.0928 \\\text { 63-under 64 } & 322 & 0.1136 \\\text { 64-under 65 } & 388 & 0.1369 \\\text { 65-under 66 } & 406 & 0.1432 \\\text { 66-under 67 } & 361 & 0.1273 \\\text { 67-under 68 } & 292 & 0.1030 \\\text { 68-under 69 } & 201 & 0.0709 \\\text { 69-under 70 } & 113 & 0.0399 \\\text { 70-under 71 } & 75 & 0.0265 \\\text { 71-under 72 } & 25 & 0.0088 \\\text { 72-under 73 } & 16 & 0.0056 \\\text { 73-under 74 } & 2 & 0.0007\end{array}$

A) $0.1831$
B) $0.4843$
C) $0.5108$
D) $0.6212$