Answer:
60.8%
Explanation:
• The mean of the women's weight = 160.3 lb
,• The standard deviation = 46.2 lb
To find the percentage of women within those limits, first, find the z-scores for x-values: 135.3 lb and 220 lb.
[tex]\begin{gathered} Z-\text{Score}=\frac{X-\text{Mean}}{S\mathrm{}D} \\ At\text{ X=}135.3,Z-\text{Score}=\frac{135.3-\text{1}60.3}{46.2}=-0.5411 \\ At\text{ X=}220,Z-\text{Score}=\frac{220-\text{1}60.3}{46.2}=1.2922 \end{gathered}[/tex]Next, using the z-score table, we find the required probability:
[tex]P(-0.5411• Therefore, the percentage of women that have weights between those limits is 60.8%.,• (b) No. A smaller percentage (39.2%) of women are excluded with those specifications.