Question

A normal distribution has a mean of μ = 100 with LaTeX: \sigmaσ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 100 and X = 130?

Answer 1

Answer:

$P(100$

And we can find this probability with this difference:

$P(0$

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

$P(0$

Step-by-step 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:

$X \sim N(100,20)$  

Where $\mu=100$ and $\sigma=20$

We are interested on this probability

$P(100$

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

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

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

$P(100$

And we can find this probability with this difference:

$P(0$

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

$P(0$

Answer 2

Answer:

Probability that the score will have a value between X = 100 and X = 130 is 0.4332.

Step-by-step explanation:

We are given that a normal distribution has a mean of μ = 100 with σ = 20 and one score is randomly selected from this distribution

Firstly, Let X = a random variable

The z score probability distribution for is given by;

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

where, $\mu$ = population mean = 100

            $\sigma$ = standard deviation = 20

Probability that the score will have a value between X = 100 and X = 130 is given by = P(100 < X < 130) = P(X < 130) - P(X $\leq$ 100)

 P(X < 130) = P( $\frac{ X - \mu}{\sigma}$ < $\frac{ 130 - 100}{20}$ ) = P(Z < 1.5) = 0.9332

 P(X $\leq$ 100) = P( $\frac{ X - \mu}{\sigma}$ $\leq$ $\frac{ 100 - 100}{20}$) = P(Z $\leq$ 0) = 0.5

Therefore, P(100 < X < 130) = 0.9332 - 0.5 = 0.4332

Hence, probability that the score will have a value between X = 100 and X = 130 is 0.4332.