Question

A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 1.25 days with an approximately normal distribution. (a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places. .10756 (b) Find the probability that their hospital stay is greater than 6 days, rounded to five decimal places.

Answer 1

Answer:

a) $P(5$

And we can find this probability with this difference:

$P(-1.90$

And using the norma standard distribution or excel we got:

$P(-1.90$

b) $P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)$

And using the complement rule we got:

$P(Z>-1.096) =1-P(Z$

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".  

Part a

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:

$X \sim N(7.37,1.25)$  

Where $\mu=7.37$ and $\sigma=1.25$

We are interested on this probability

$P(5$

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:

$P(5$

And we can find this probability with this difference:

$P(-1.90$

And using the norma standard distribution or excel we got:

$P(-1.90$

Part b

For this case we want this probability:

$P(X>6)$

And we can use the z score and we got:

$P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)$

And using the complement rule we got:

$P(Z>-1.096) =1-P(Z$