Question

Suppose the number of hours to complete this exam for my students is denoted by the random variable X and is normally distributed with the mean of µx hours and the standard deviation of σX = 2.3 hours. Find the probability that a randomly selected student has finished the exam spending 3 to 6 hours, P(3 < X < 6) =? (Use the attached table.)

Answer 1

Answer:

$P(3$

And we can find this probability with this difference:

$P(-0.65$

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

$P(-0.65$

Step-by-step explanation:

Asuming this complete question: Suppose the number of hours to complete this exam for my students is denoted by the random variable X and is normally distributed with the mean of µx hours and the standard deviation of σX = 2.3 hours. Find the probability that a randomly selected student has finished the exam spending 3 to 6 hours, P(3 < X < 6) =? (Use the attached table.)

Let =4.5

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 number of hours to complete an examn of a population, and for this case we know the distribution for X is given by:

$X \sim N(4.5,2.3)$  

Where $\mu=4.5$ and $\sigma=2.3$

We are interested on this probability

$P(3$

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:

$P(3$

And we can find this probability with this difference:

$P(-0.65$

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

$P(-0.65$