Question

Adult male lion weights are normally distributed, with a mean of 420 pounds and a standard deviation of 17 pounds. Approximately what percentage of the adult lions weigh more than 450 pounds?
96.08%
97.22%
1.76%
3.92%

Answer 1

Answer:

3.92%

Step-by-step explanation:

First calculate Z:

$Z= (x - mean)/(st. dev.) \\Z = (450 - 420)/17 \\Z= 1.76$

The percentage for a Z of 1.76 is 96.08%.

That means that all lions up to Z = 1.76 are equivalent to 96.08% but you are searching for the lions above Z = 1.76, then 100 - 96.08 = 3.92%

Answer 2

Answer:

$P(X>450)=P(\frac{X-\mu}{\sigma}>\frac{450-\mu}{\sigma})=P(Z>\frac{450-420}{17})=P(z>1.765)$

And we can find this probability with the complement rule and with excel or the normal standard distribution:

$P(z>1.765)=1-P(z$

And that correspond to :

3.92%

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 weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(420,17)$  

Where $\mu=420$ and $\sigma=17$

We are interested on this probability

$P(X>450)$

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:

$P(X>450)=P(\frac{X-\mu}{\sigma}>\frac{450-\mu}{\sigma})=P(Z>\frac{450-420}{17})=P(z>1.765)$

And we can find this probability with the complement rule and with excel or the normal standard distribution:

$P(z>1.765)=1-P(z$

And that correspond to :

3.92%