Question

in a recent study in london, it was found the systolic blood pressure of 36 - year - old females followed an approximately normal distribution with a mean of 117 mmhg and a standard deviation of 15 mmhg.
a. label the values for the mean, along with the values for 1, 2, and 3 standard deviations away from the mean.
b. what percent of women in the study had a systolic blood pressure between 87 and 147 mmhg?
c. what percent of women in the study had a systolic blood pressure higher than 132 mmhg?
d. what percent of women in the study had a systolic blood pressure between 87 and 132 mmhg?
e. how many standard deviations away from the mean is a systolic blood pressure of 135?

Explanation:

Step1: Calculate values for standard - deviations

Given mean $\mu = 117$ and standard deviation $\sigma=15$.
1 standard deviation above mean: $\mu+\sigma=117 + 15=132$; 1 standard deviation below mean: $\mu - \sigma=117-15 = 102$; 2 standard deviations above mean: $\mu + 2\sigma=117+2\times15=147$; 2 standard deviations below mean: $\mu-2\sigma=117 - 2\times15=87$; 3 standard deviations above mean: $\mu+3\sigma=117+3\times15 = 162$; 3 standard deviations below mean: $\mu - 3\sigma=117-3\times15=72$.

Step2: Use the empirical rule for part b

The empirical rule for a normal distribution states that approximately 99.7% of the data lies within 3 standard - deviations of the mean. The values 87 and 147 are 2 standard deviations below and above the mean respectively. So the percentage of women with systolic blood pressure between 87 and 147 mmHg is approximately 95%.

Step3: Use the empirical rule for part c

The value 132 is 1 standard deviation above the mean. The percentage of data above 1 standard deviation above the mean is $\frac{100 - 68}{2}=16\%$.

Step4: Use the empirical rule for part d

The value 87 is 2 standard deviations below the mean and 132 is 1 standard deviation above the mean. The percentage of data between 2 standard deviations below and 1 standard deviation above the mean is $\frac{95+68}{2}=81.5\%$.

Step5: Calculate z - score for part e

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Substitute $x = 135$, $\mu=117$, and $\sigma = 15$ into the formula: $z=\frac{135 - 117}{15}=\frac{18}{15}=1.2$. So 135 is 1.2 standard deviations away from the mean.

Answer:

a. Mean: 117; 1 standard deviation above mean: 132, 1 standard deviation below mean: 102; 2 standard deviations above mean: 147, 2 standard deviations below mean: 87; 3 standard deviations above mean: 162, 3 standard deviations below mean: 72.
b. 95%
c. 16%
d. 81.5%
e. 1.2 standard deviations