# Introduction to Normal Distribution #2

We learned about the normal distribution in the first article. We will solve the practice problems shown in the first article to familiarize ourselves with this distribution

## Empirical Rule problem

Find out the percentage of the following data from above example with mean 60 and standard deviation 15 and sample size 10,000.

What percentage of shells have size below 75mm ?
When looking at the empirical rule of normal distribution we see that 75 is at the border between Yellow and Blue region to the Right of the mean, which is 60. So if we add all the blocks that are to the LEFT of 75 then we will get the answer.
%(X<75) = 0.135 + 2.14 + 13.59 + 34.135 + 34.135 = 84.135%

What percentage of shells have size between 15 and 90mm ?
For This we need to add the regions between 15 and 90.
% (15<X<90) = 2.14 + 13.59 + 34.135 + 34.135 + 13.59 = 97.59%

What percentage of shells have size above 60mm ?
This question may seem tricky, but we know that 60 is the mean. That tells us that 50% of shells will have more than 60mm size according to the distribution.

How many shells have a size between 30 and 60mm ?
The answer to this question is the sum of the blue and yellow region between 30 and 60
%(30<X<60) = 13.59 + 34.135 = 47.725%

How many shells have size between 45 and 105mm ?
We add the regions between 45 and 105 to get the answer for this question
%(45<X<105) = 34.135 + 34.135 + 13.59 + 2.14 = 84%

How many shells have size less than 30mm?
We add the regions to the LEFT of 30 to get the answer
%(X<30) = 0.135 + 2.14 = 2.275%

## Standard Normal Distribution problem

A small city has a population of n=50,000 with average age of 40 with a standard deviation of 10. Answer the following questions. (Hint Use Z Table to solve the problems)

What is the probability that a random person you meet has age less than 42?
We will calculate the Z score formula.
Z = (42 – 40)/10 = 2/10 = 0.2
We have a positive Z value so we will use the Positive Z table and get the P value
P (X < 42) = 0.5793

What is the probability that a random person chosen has age between 33 and 64?
We will calculate 2 Z scores and 2 P values to get the answer for our problem
Z1 = (64 – 40)/10 = 24/10 = 2.4
Z2 = (33 – 40)/10 = -7/10 = -0.7
We will get P1 from Positive Z table and P2 from negative Z table.
P1(X1 < 64) = 0.9918
P2(X2 < 33) = 0.2420
Both values of P1 and P2 are cumulative areas under the standard normal distribution to the LEFT of the Z values. Thus we get area between the 2 X values when we subtract P2 from P1
P(33<X<64) = P1(X1 < 64) – P2(X2 < 33) = 0.9918 – 0.2420 = 0.7498

How many people are expected to have age between 25 and 55?
We will calculate 2 Z scores and 2 P values to get the answer for our problem
Z1 = (55 – 40)/10 = 15/10 = 1.5
Z2 = (25 – 40)/10 = -15/10 = -1.5
We will get P1 from Positive Z table and P2 from negative Z table.
P1(X1 < 55) = 0.9332
P2(X2 < 25) = 0.0668
Both values of P1 and P2 are cumulative areas under the standard normal distribution to the LEFT of the Z values. Thus we get area between the 2 X values when we subtract P2 from P1
P(25<X<55) = P1(X1 < 55) – P2(X2 < 25) = 0.9332 – 0.0668 = 0.8664

How many people in the town may have age above 75?
We will calculate the Z score formula.
Z = (75 – 40)/10 = 35/10 = 3.5
Using the Z table we will get the P value
P (X<75) = 0.9998
Note: The P value we have calculated is the area under the curve that is LESS than X = 75.
The problem question asks us for age ABOVE 75. So we use the complementary property of Probability to get the correct P value
P (X > 75) = 1 – P(X<75) = 1 – 0.9998 = 0.0002
Now that we have the probability of a person above the age of 75 we can calculate the expected number of people above 75 in the town by the formula
N>75 = n*p = 50,000*0.0002 = 10 people
Where n = town population and p = calculated probability.

What is the probability that a group of 50 people in a restaurant will have ages between 36 and 45?
We are given a sample size of 50 in this question. so we will use the Z score formula shown on the right side.
We will calculate 2 Z scores and 2 P values to get the answer for our problem
Z1 = (45 – 40)/(10/sqrt50) = 5/1.4142 = 3.53
Z2 = (36 – 40)/(10/sqrt50) = -4/1.4142 = -2.83
We will get P1 from Positive Z table and P2 from negative Z table.
P1(X1 < 45) = 0.9998
P2(X2 < 36) = 0.0023
Both values of P1 and P2 are cumulative areas under the standard normal distribution to the LEFT of the Z values. Thus we get area between the 2 X values when we subtract P2 from P1
P(36<X<45) = P1(X1 < 45) – P2(X2 < 36) = 0.9998 – 0.0023 = 0.9975

Try our Normal distribution Quiz for more practice!