Question

Assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. Approximately what percentage of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ microliter?

Answer 1

Answer:

82.31% of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ micro liter.

Step-by-step explanation:

We are given that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per micro liter and a standard deviation of 0.382 million cells per micro liter.

Let X = red blood cell counts of women

The z-score probability distribution for is given by;

               Z = $\frac{ X -\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean count = 4.577 million cells per micro liter

            $\sigma$ = standard deviation = 0.382 million cells per micro liter

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of women having red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ micro liter is given by = P(4.2 < X < 5.4) = P(X < 5.4 million) - P(X $\leq$ 4.2 million)

  P(X < 5.4) = P( $\frac{ X -\mu}{\sigma}$ < $\frac{5.4-4.577}{0.382}$ ) = P(Z < 2.15) = 0.98422  {using z table}

  P(X $\leq$ 4.2) = P( $\frac{ X -\mu}{\sigma}$ $\leq$ $\frac{4.2-4.577}{0.382}$ ) = P(Z $\leq$ -0.99) = 1 - P(Z < 0.99)

                                                     = 1 - 0.83891 = 0.16109

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.15 and x = 0.99 in the z table which has an area of 0.98422 and 0.83891 respectively.

Therefore, P(4.2 < X < 5.4) = 0.98422 - 0.16109 = 0.8231 or 82.31%

Hence, 82.31% of the women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ micro liter.

Answer 2

Answer:

82.31% of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ microliter

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 4.577, \sigma = 0.382$

Approximately what percentage of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ microliter?

This is the pvalue of Z when X = 5.4 subtracted by the pvalue of Z when X = 4.2. So

X = 5.4

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.4 - 4.577}{0.382}$

$Z = 2.15$

$Z = 2.15$ has a pvalue of 0.9842

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 4.577}{0.382}$

$Z = -0.99$

$Z = -0.99$ has a pvalue of 0.1611

0.9842 - 0.1611 = 0.8231

82.31%  of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per​ microliter