To answer the question, having a z-table with you will help. We can also use the 68-95-99.7 rule.
The rule states that 68.27% of a normally distributed data set is within one standard deviation of the mean, 95.45% is within two standard deviations, and 99.73% is with three standard deviations.
Given that the mean is 131 mmHg and the standard deviation is 9 mmHg, we can calculate the boundaries which are 3 standard deviations away from the mean by adding and subtracting three times the standard deviation.
[tex]\begin{gathered} 131-(3\times9)=104 \\ \\ 131+(3\times9)=158 \end{gathered}[/tex]
Therefore, approximately 99.7% of women over seventy have blood pressures between 104 mmHg and 158 mmHg.
Now let's find out how many standard deviations away 122 mmHg and 140 mmHg are from the mean.
[tex]\begin{gathered} z=\frac{122-131}{9}=-1 \\ \\ z=\frac{140-131}{9}=1 \end{gathered}[/tex]
122 and 140 mmHg are within 1 standard deviation of the mean. Using the 68-95-99.7 rule, we know that approximately 68.27% of women over seventy have blood pressures between 122 mmHg and 140 mmHg.
