Question

Find the following percentiles for the standard normal distribution. Interpolate where appropriate. (Round your answers to two decimal places.)

(a)    81st


(b)    19th


(c)    76th


(d)    24th


(e)    10th

Answer 1

Concepts and reason

A statistical measure used to indicate the value below certain percentage of observations in a specified group is called percentile.

For example, the 50th percentile is the middle value of the group which is under study. In other words the value represents the median value.

Fundamentals

The $\alpha$ % percentile is defined as follows:

$P\left( {X \le x} \right) = \alpha \%$

The excel function to compute the percentile value is ${\rm{ = NORMSINV(probability)}}$

(a)

The mean and standard deviation of the standard normal distribution is $\mu = 0$ and standard deviation $\sigma = 1$

The 81^st percentile is calculated as follows:

$\begin{array}{c}\\{p_{81}} = NORMSINV\left( {0.81} \right)\\\\ = 0.877896\\\\ \approx 0.88\\\end{array}$

(b)

The mean and standard deviation of the standard normal distribution is $\mu = 0$ and standard deviation $\sigma = 1$

The 19^th percentile is calculated as follows:

$\begin{array}{c}\\{p_{19}} = NORMSINV\left( {0.19} \right)\\\\ = - 0.8779\\\\ \approx - 0.88\\\end{array}$

(c)

The mean and standard deviation of the standard normal distribution is $\mu = 0$ and standard deviation $\sigma = 1$

The 76^th percentile is calculated as follows:

$\begin{array}{c}\\{p_{76}} = NORMSINV\left( {0.76} \right)\\\\ = 0.706303\\\\ \approx 0.71\\\end{array}$

(d)

The mean and standard deviation of the standard normal distribution is $\mu = 0$ and standard deviation $\sigma = 1$

The 24^th percentile is calculated as follows:

$\begin{array}{c}\\{p_{24}} = NORMSINV\left( {0.24} \right)\\\\ = - 0.706303\\\\ \approx - 0.71\\\end{array}$

(e)

The mean and standard deviation of the standard normal distribution is $\mu = 0$ and standard deviation $\sigma = 1$

The 10^th percentile is calculated as follows:

$\begin{array}{c}\\{p_{10}} = NORMSINV\left( {0.10} \right)\\\\ = - 1.28155\\\\ \approx - 1.28\\\end{array}$

Ans: Part a

The 81^st percentile for the standard normal distribution is 0.88.

Part b

The 19^th percentile for the standard normal distribution is -0.88.

Part c

The 76^th percentile for the standard normal distribution is 0.71.

Part d

The 24^th percentile for the standard normal distribution is -0.71.

Part e

The 10^th percentile for the standard normal distribution is -1.28.