Question

The average number of cocktails that residents of my nursing home drink weekly is normally distributed with mean of 12 cocktails with a standard deviation of 3 cocktails. What is the percentile rank of a resident who drinks 11.2 cocktails weekly?

about 32% about 11% about 61% about 5% about 95% about 18% about 39% about 82%

Answer 1

According to the question, the average number of cocktails that residents of my nursing home drink weekly is normally distributed with mean of 12 cocktails with a standard deviation of 3 cocktails.

Let us define, X : the no. of cocktails that residents of my nursing home drink weekly,

Then,

i.e. Z follows a standard normal distribution.

Now, the percentile rank of an observation/score is the percentage of observations/scores in a frequency distribution that are equal or lower than it.

We are to find the percentile rank of a resident who drinks 11.2 cocktails weekly. Also, we know that cumulative probability of a particular observation of standard normal variate is given by

, where k is the observed value of Z.

Then,

[Since the distribution of Z(i.e. a standard normal variate) is symmetrical about 0]

[Obtained from the Table of ordinates and area of Standard Normal variable]

Then the relation between percentile rank & cumulative probability of a standard normal variate can be approximated as:

percentile rank= (cumulative probability * 100)

Therefore,

percentile rank of a resident who drinks 11.2 cocktails weekly

(approximately)

So. the percentile rank of a resident who drinks 11.2 cocktails weekly is about 39%.