Question

The red blood cell counts​ (in millions of cells per​ microliter) for a population of adult males can be approximated by a normal​ distribution, with a mean of 5.8 million cells per microliter and a standard deviation of 0.4 million cells per microliter.

(a) What is the minimum red blood cell count that can be in the top 21 %of counts?

(b) What is the maximum red blood cell count that can be in the bottom 11 %of counts?

Answer 1

Answer:

(a) The minimum red blood cell count that can be in the top 21% of counts is 6.1.

(b) The maximum red blood cell count that can be in the bottom 11% of counts is 5.3.

Step-by-step explanation:

The random variable X can be defined as the number of red blood cells in adult males.

The random variable X can be approximated by a normal​ distribution.

The mean number of red blood cells is, μ = 5.8 million cells per micro liter.

The standard deviation of red blood cells is, σ = 0.4 million cells per micro liter.

(a)

Compute the value of x such that P (X > x) = 0.21 as follows:

   P (X > x) = 0.21

⇒ P (Z > z) = 0.21

The value of z is, 0.81.

*Use a z-table.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\0.81=\frac{x-5.8}{0.4}\\x=5.8+(0.4\times 0.81)\\x=6.124\\\approx6.1$

Thus, the minimum red blood cell count that can be in the top 21% of counts is 6.1.

(b)

Compute the value of x such that P (X < x) = 0.11 as follows:

   P (X < x) = 0.11

⇒ P (Z < z) = 0.11

The value of z is, -1.23.

*Use a z-table.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\-1.23=\frac{x-5.8}{0.4}\\x=5.8-(0.4\times 1.23)\\x=5.308\\\approx5.3$

Thus, the maximum red blood cell count that can be in the bottom 11% of counts is 5.3.

Answer 2

Using the normal distribution, we have that the blood counts are:

a) 6.12 million cells per microliter.

b) 4.81 million cells per microliter.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• Mean of 5.8 million cells per microliter, hence $\mu = 5.8$
• Standard deviation of 0.4 million cells per microliter, hence $\sigma = 0.4$

Item a:

This is the 100 - 21 = 79th percentile, which is X when Z has a p-value of 0.79, so X when Z = 0.807.

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

$0.807 = \frac{X - 5.8}{0.4}$

$X - 5.8 = 0.4(0.807)$

$X = 6.12$

Hence, 6.12 million cells per microliter.

Item b:

This is the 11th percentile, which is X when Z has a p-value of 0.11, so X when Z = -1.227.

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

$-1.227 = \frac{X - 5.8}{0.4}$

$X - 5.8 = -1.227(0.807)$

$X = 4.81$

Hence, 4.81 million cells per microliter.

A similar problem is given at https://brainly.com/question/24663213