Question

For women aged 18-24, systolic blood pressure (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. If 23 women aged 18-24 are randomly selected, find the probability that their mean systolic blood pressure is between 119 and 122.

Answer 1

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Answer 2

Concepts and reason

The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean and any particular normal distribution is completely specified by the numbers mean and standard deviation.

Fundamentals

The formula for standard error is, $SE = \frac{\sigma }{{\sqrt n }}$

The formula of standard normal score is, $Z = \frac{{\bar X - \mu }}{{SE}}$

Here, $\mu$ is the population mean and $\sigma$ is the population standard deviation of the normal distribution.

Calculate the standard error value of women aged 18-24 systolic blood pressure.

From the given information: $n = 23;\mu = 114.8{\rm{ and }}\sigma = 13.1$

$\begin{array}{c}\\SE = \frac{\sigma }{{\sqrt n }}\\\\ = \frac{{13.1}}{{\sqrt {23} }}\\\\ = {\rm{2}}.{\rm{731539}}\\\\ = 2.732{\rm{ }}\left( {{\rm{Rounded to 3 decimal place}}} \right)\\\end{array}$

Calculate the probability that women aged 18-24 systolic blood pressure is between 119 and 122.

Let $\bar X$ denote the women aged 18-24 systolic blood pressure.

$\begin{array}{c}\\P\left( {119 < \bar X < 122} \right) = P\left( {\frac{{119 - 114.8}}{{2.732}} < \frac{{\bar X - \mu }}{{SE}} < \frac{{122 - 114.8}}{{2.732}}} \right)\\\\ = P\left( {1.54 < Z < 2.64} \right)\\\\ = P\left( {Z < 2.64} \right) - P\left( {Z < 1.54} \right)\\\end{array}$

$\begin{array}{c}\\ = \left[ \begin{array}{l}\\\left( {{\rm{ = NORMSDIST(2}}{\rm{.64}}} \right) - \\\\\left( {{\rm{ = NORMSDIST(1}}{\rm{.54}}} \right)\\\end{array} \right]{\rm{ }}\left( {{\rm{Use MS Excel}}} \right)\\\\ = 0.9959 - 0.9382\\\\ = 0.0577\\\end{array}$

Ans:

The probability that women aged 18-24 systolic blood pressure is between 119 and 122 is: 0.0577.