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Solve the Problem A)04591
B)0

Question 20

Multiple Choice

Solve the problem.
-A southeastern college has an enrollment of 2951 female students. Records show that the mean height of these students is 64.7 inches and that the standard deviation is 2.3 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 µ = 64.7 and Ϭ = 2.3) . Making
This assumption, estimate the area under the associated normal curve to the left of 65.  Height  (inches)   Freq.  Relative  freq.  56-under 57 20.0007 57-under 58 70.0024 58-under 59 140.0047 59-under 60 310.0105 60-under 61 1010.0342 61-under 62 1940.0657 62-under 63 3110.1054 63-under 64 4100.1389 64-under 65 5260.1782 65-under 66 4820.1633 66-under 67 3970.1345 67-under 68 2540.0861 68-under 69 1500.0508 69-under 70 490.0166 70-under 71 170.0058 71-under 72 50.0017 72-under 73 10.0003\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 { 56-under 57 } & 2 & 0.0007 \\\text { 57-under 58 } & 7 & 0.0024 \\\text { 58-under 59 } & 14 & 0.0047 \\\text { 59-under 60 } & 31 & 0.0105 \\\text { 60-under 61 } & 101 & 0.0342 \\\text { 61-under 62 } & 194 & 0.0657 \\\text { 62-under 63 } & 311 & 0.1054 \\\text { 63-under 64 } & 410 & 0.1389 \\\text { 64-under 65 } & 526 & 0.1782 \\\text { 65-under 66 } & 482 & 0.1633 \\\text { 66-under 67 } & 397 & 0.1345 \\\text { 67-under 68 } & 254 & 0.0861 \\\text { 68-under 69 } & 150 & 0.0508 \\\text { 69-under 70 } & 49 & 0.0166 \\\text { 70-under 71 } & 17 & 0.0058 \\\text { 71-under 72 } & 5 & 0.0017 \\\text { 72-under 73 } & 1 & 0.0003\end{array}


A) 0.4591
B) 0.5407
C) 0.3415
D) 0.1782

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