Mathematics, 28.07.2019 21:30, chloeboo

The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 248.5 and a standard deviation of 61.1. (all units are 1000 cells/mul.) using the empirical rule, find each approximate percentage below. a. what is the approximate percentage of women with platelet counts within 2 standard deviations of the mean, or between 126.3 and 370.7? b. what is the approximate percentage of women with platelet counts between 65.2 and 431.8?

Answers: 2

Show answers

Answers

answered: Roof55

The blood platelet count is an illustration of normal distribution

Approximately 95% of the data lies within 2 standard deviations of the mean.There are approximately 99.7% of women with platelet count between 65.2 and 431.8

The given parameters are:

$\mathbf{\mu = 248.5}$

$\mathbf{\sigma = 61.1}$

(a) The percentage within 2 standard deviation of mean or between 126.3 and 370.7

Start by calculating the z-score, when x = 126.3 and x = 370.7

$\mathbf{z = \frac{x - \mu}{\sigma}}$

So, we have:

$\mathbf{z = \frac{126.3 -248.5}{61.1}}$

$\mathbf{z = \frac{-122.2}{61.1}}$

$\mathbf{z = -2}$

Also:

$\mathbf{z = \frac{370.7 -248.5}{61.1}}$

$\mathbf{z = \frac{122.2}{61.1}}$

$\mathbf{z = 2}$

The empirical rule states that:

Approximately 95% of the data lies within 2 standard deviations of the mean.

Hence, there are approximately 95% of women with platelet count within 2 standard deviations of the mean.

(b) The percentage with platelet count between 65.2 and 431.8

Start by calculating the z-score, when x = 65.2 and x = 431.8

$\mathbf{z = \frac{x - \mu}{\sigma}}$

So, we have:

$\mathbf{z = \frac{65.2 -248.5}{61.1}}$

$\mathbf{z = \frac{-183.3}{61.1}}$

$\mathbf{z = -3}$

Also:

$\mathbf{z = \frac{431.8 -248.5}{61.1}}$

$\mathbf{z = \frac{183.3}{61.1}}$

$\mathbf{z = 3}$

The empirical rule states that:

Approximately 99.7% of the data lies within 3 standard deviations of the mean.

Hence, there are approximately 99.7% of women with platelet count between 65.2 and 431.8

Read more about normal distributions at:

link

answered: damariush567

a. Approximately 95% women have platelet counts between 126.3 and 370.7.

b. Approximately 99.7% women have platelet counts between 65.2 and 431.8.

Step-by-step explanation:

We have been given that the blood platelet counts of a group of women have a bell-shaped distribution with a mean of 248.5 and a standard deviation of 61.1.

a. Empirical rule of normal distribution states:

$\approx68\%$ of the data lies within 1 standard deviation of the mean

$\approx95\%$ of the data lies within 2 standard deviations of the mean

$\approx99.7\%$ of the data lies within 3 standard deviations of the mean.

Now let us find z-score for our given values as z-score represents that a data point is how many standard deviation away from mean.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{126.3-248.5}{61.1}$

$z=\frac{-122.2}{61.1}$

z=-2

$z=\frac{370.7-248.5}{61.1}$

$z=\frac{122.2}{61.1}$

z=2

Since our given values are -2 and 2 standard deviations of mean, therefore, approximately 95% of women have platelet counts between 126.3 and 370.7.

b. Let us find z-score for our given values as z-score represents that a data point is how many standard deviation away from mean.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{65.2-248.5}{61.1}$

$z=\frac{-183.3}{61.1}$

z=-3

$z=\frac{431.8-248.5}{61.1}$

$z=\frac{183.3}{61.1}$

z=3

Since our given values are -3 and 3 standard deviations of mean, therefore, approximately 99.7% of women have platelet counts between 65.2 and 431.8.

answered: Guest

the answer is. -82562234434584

step-by-step explanation:

answered: Guest

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ a(\stackrel{x_1}{5}~,~\stackrel{y_1}{4})\qquad b(\stackrel{x_2}{x}~,~\stackrel{y_2}{y})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{ab}{7}=\sqrt{(x-5)^2+(y-4)^2}~\hfill$