Question

Normal Distribution Problem. Red Blood Cell Counts are expressed millions per cubic millimeter of whole blood. For healthy females, x has a approximately normal distribution with mu = 4.8 and sigma =.3. Note: my probabilities are exact probabilities. Solutions using the standard normal table will be close.What is the probability that a healthy female has a red blood count between 3.9 and 5.0?a..4787b..2475c..7248d. .7462

Answer 1

Answer: option d is the correct answer.

Explanation:

For healthy females, x has a approximately normal distribution. We would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = red blood counts.

µ = mean count

σ = standard deviation

From the information given,

µ = 4.8

σ = 0.3

We want to find the that a healthy female has a red blood count between 3.9 and 5.0. It is expressed as

P(3.9 ≤ x ≤ 5) = P(x ≤ 5) - P(x ≤ 3.9)

For x = 3.9,

z = (3.9 - 4.8)/0.3 = - 3

Looking at the normal distribution table, the probability corresponding to the z score is 0.00135

For x = 5,

z = (5 - 4.8)/0.3 = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.7486

P(3.9 ≤ x ≤ 5) = = 0.7486 - 0.00135

= 0.747

Answer 2

`P(3.9<X<5)=P((3.9-\mu)/(\sigma)<(X-\mu)/(\sigma)<(5-\mu)/(\sigma))=P((3.9-4.8)/(0.3)<Z<(5-4.8)/(0.3))=P(-3<z<0.67)`

`P(-3<z<0.67)=P(z<0.67)-P(z<-3)`

`P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722`

d. .7462

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

`X \sim N(4.8,0.3)`

Where
`\mu=4.8´` and
`\sigma=0.3`

We are interested on this probability

`P(3.9<X<5.0)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(3.9<X<5)=P((3.9-\mu)/(\sigma)<(X-\mu)/(\sigma)<(5-\mu)/(\sigma))=P((3.9-4.8)/(0.3)<Z<(5-4.8)/(0.3))=P(-3<z<0.67)`

And we can find this probability with this difference:

`P(-3<z<0.67)=P(z<0.67)-P(z<-3)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722`

And on this case the most accurate answer would be:

d. .7462