Question

Weights of male mountain lions follow the normal distribution with a median of 150 lb and an interquartile range of 8.2 lb, What is the 95th percentile of the weights?

Answer 1

The standard deviation can be calculated using the formula
\[\sigma=\frac{interquartile\ range}{1.34896}\]
In this case the standard deviation 8.2/1.34896= 6.079
A standard normal distribution table shows the z-score for a cumulative probability of 95% is
z = 1.645
The median and the mean are the same. Therefore the following formula can be used:
\[z=\frac{X-\mu}{\sigma}\]
Substituting we get
\[1.645=\frac{X-150}{6.079}\]
Solving for X gives
X = 150 + 9.9999

Answer 2

Thank you; I am still perusing it to understand the workings. How did you come about the formulas and all that? Could you explain a little more please? And how is the mean and median the same?

Answer 3

The mean and the median are the same in a normal distribution.

The IQR (the width of an interval which contains the middle 50% of the data set) is normally computed by subtracting the first quartile from the third quartile. In a normal distribution (with mean 0 and standard deviation 1), the first and third quartiles are located at -0.67448 and +0.67448 respectively. Thus the IQR for a normal distribution is:
\[IQR=Q _{3}-Q _{1}=2(0.67448)\times \sigma=1.34896 \sigma\]
Interquartile range = 1.34896 x standard deviation

Answer 4

Thank you so much for the explanation; I appreciate it.

Answer 5

You're welcome :)