Question

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

standard deviation is 1.5 years.

Use the empirical rule (68 – 95 - 99.7%) to estimate the probability of a meerkat living less than 14.6

years.

Answer 1

Answer:

$P(x<14.6)$

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:

$P(X<14.6) = 0.84$

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:

$P(x<14.6)$

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:

$P(X<14.6) = 0.84$

Answer 2

Using the Empirical Rule, it is found that there is a 84% probability of a meerkat living less than 14.6.

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• The Empirical Rule states that in a normal distribution, 68% of the measures are within 1 standard deviation of the mean, 95% are within 2 standard deviations and 99.7% are within 3 standard deviations.

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• The mean is of 13.1 years, while the standard deviation is of 1.5 years.
• We have also consider that the normal distribution is symmetric, which means that 50% of the measures are above the mean and 50% are below.
• 14.6 = 13.1 + 1.5, thus, one standard deviation above the mean.
• Of the 50% of the measures below the mean, all are below 14.6, while of the 50% above, 68% are below 14.6, thus:

$P = 50 + 0.68(50) = 50 + 34 = 84$

84% probability of a meerkat living less than 14.6.

A similar problem is given at https://brainly.com/question/13503878