Question

The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 13.1 years; the

Answer 1

Answer: 16, if the question asks for longer than 14.6 years

Step-by-step explanation:

Answer 2

Answer:

$P(x$

We can calculate the z score given by:

$z =\frac{x-\mu}{\sigma}= \frac{14.6-13.1}{1.5}= 1$

So for this case 14.6 is a value 1 deviation above the mean

We know that within 1 deviation from the mean we have 68% of the data, and on the tails we will have 100-68 = 32% and 16% on each tail, so then we can conclude that:

$P(X$

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the lifespans of meerkats.

From the problem we have the mean and the standard deviation for the random variable X. $E(X)=13.1, Sd(X)=1.5$

So we can assume $\mu=13.1 , \sigma=1.5$

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

And we want the following probability:

$P(x$

We can calculate the z score given by:

$z =\frac{x-\mu}{\sigma}= \frac{14.6-13.1}{1.5}= 1$

So for this case 14.6 is a value 1 deviation above the mean

We know that within 1 deviation from the mean we have 68% of the data, and on the tails we will have 100-68 = 32% and 16% on each tail, so then we can conclude that:

$P(X$