Question

The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives 22.4 years; the

Answer 1

Answer:

$P(27.8 < X 27.8)-P(X>30.5)=0.025-0.0015=0.0235$

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 tigers in a particular zoo.

From the problem we have the mean and the standard deviation for the random variable X. $\mu=120 , \sigma=110$

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

We want to find this probability:

$P(27.8 < X$

And in order to calculate how many deviation we are above/below the mean we can use the z score given by:

$z =\frac{x-\mu}{\sigma}$

And if we use this formula for the two values given we have:

$z_1 = \frac{27.8-22.4}{2.7}=2$

$z_1 = \frac{30.5-22.4}{2.7}=3$

So we have values between 2 and 3 deviations above the mean.

We can use the following probabilities

$P(X$    

$P(X>\mu +\sigma)=P(X >25.1)=0.16$  

$P(X$    

$P(X>\mu +2*\sigma)=P(X>27.8)=0.025$

$P(X$

$P(X>\mu +3*\sigma)=P(X>30.5)=0.0015$

And we can find this probability on this way:

$P(27.8 < X 27.8)-P(X>30.5)=0.025-0.0015=0.0235$