# Normal Distribution/Gaussian Distribution

Normal distribution is a continuous probability distribution.Its graph is called normal curve.

Johann Carl Friedrich Gauss, a German mathematician, played a major role in deriving the Normal distribution, and hence, normal distribution is also known Gaussian distribution.

A continuous random variable *x* is said to follow Normal distribution if it has a probability density function

Where Î¼ is the mean of the distribution and is the variance.

A normal variate with parameters Î¼ and Ïƒ is denoted by *N*(Î¼, Ïƒ).

# Properties of Normal Distribution

- It is a bell-shaped curve.
- The distribution is symmetrical about the mean Î¼.
- Since the curve is symmetric about Î¼, the mean, mode and median coincide,
*i.e.*, - It is unimodal distribution (single mode).
- The standard deviation for a normal distribution is given by Ïƒ.
- The mean deviation is MD = 4/5Ïƒ. The quartile deviation is Q.D = 2/3Ïƒ
- The quartiles are given by
- Lower quartile
- Upper quartile

- So the value of quartile deviation is 0.675 Ïƒ.
- (Î¼ + Ïƒ) and (Î¼ - Ïƒ) are known as the points of inflexion,
*i.e.*, the curve changes its curvature from concave to convex and form convex to concave at the two points, respectively. - Asymptotic base: The two tails of the curve extend indefinitely and never touch the horizontal axis.
- The area under the curve between two limits is constant,
*i.e.*,- Area between is 0.6828 of the total area under the curve
- Area between is 0.9846 of the total area under the curve
- Area between is 0.9973 of the total area under the curve

# Applications of Normal Distribution

- Normal distribution can be manipulated for social purposes and natural sciences.
- Binomial distribution tends to normal distribution when the number of trials is very large and the probability of success is moderate.
- Poisson distribution also tends to normal distribution for large values of
*m*. - It is very useful in statistical quality control.

# Standard Normal Distribution

Standard normal distribution is very similar to a normal distribution only in this case, the mean, median and mode are all equal to zero and the standard deviation is equal to 1.So,

A continuous random variable *z* is said to follow normal distribution if it has a probability density function

Substituting Î¼ = 0 and Ïƒ^{2} = 1, we get the probability function for standard normal distribution for a continuous random variable *z* as

# Properties of Standard Normal Distribution

The probability curve for standard normal distribution can be obtained by plotting the function y = f(x).- As we saw earlier, mean = mode = median = 0.
- The standard deviation is
- The mean deviation is given by 0.8Ïƒ = 0.8.
- The quartile deviation is given by 0.675Ïƒ = 0.675.
- The two points of inflexion are +1 and -1.

Example

The lifetimes of certain kinds of electronic devices have a mean of 300 hours and standard deviations of 25 hours. Assuming that the distribution of these lifetimes, which are measured to the nearest hour, can be approximated closely with a normal curve,

a. find the probability that any one of these electronic devices will have a lifetime of more than 350 hours

b. find the percentage of lifetimes of these electronic devices more than 300 hours or less

c. find the percentage of lifetimes of these electronic devices will have lifetimes from 220 or 260 hours

a. find the probability that any one of these electronic devices will have a lifetime of more than 350 hours

b. find the percentage of lifetimes of these electronic devices more than 300 hours or less

c. find the percentage of lifetimes of these electronic devices will have lifetimes from 220 or 260 hours

Solution

Given

Thus, the required probability is 1 â€“ 0.9772 = 0.0228.

a. âˆ´ The required percentage is 0.5000 x 100 = 50%.

b. Given

From the normal table, we have

Thus, the required probability is

*Î¼*= 300,*Ïƒ*= 25 and*x*= 350, then The area under normal curve between*z*= 0 and*z*= 2 is 0.9772.Thus, the required probability is 1 â€“ 0.9772 = 0.0228.

a. âˆ´ The required percentage is 0.5000 x 100 = 50%.

b. Given

*x*_{1}= 220,*x*_{2}= 260,*Î¼*= 300 and*Ïƒ*= 25. Thus,From the normal table, we have

*P*(*z*= -1.6) = 0.4452 and*P*(*z*= - 3.2) = 0.4903.Thus, the required probability is

*P*(*z*= -3.2) -*P*(*z*= -1.6) = 0.4903 - 0.4452 = 0.0541.

Example

Assume that the test scores from a college admissions test are normally distributed and it should be in order with a mean of 450 and a standard deviation of 100.

A. What percentage of people taking the test score are between 400 and 500?

B. Suppose, someone received a score of 630. What percentage of people taking the test score better? What percentage score worse?

C. If particular university will not admit any one scoring below 480, what percentage of persons taking the test would be acceptable to that university?

A. What percentage of people taking the test score are between 400 and 500?

B. Suppose, someone received a score of 630. What percentage of people taking the test score better? What percentage score worse?

C. If particular university will not admit any one scoring below 480, what percentage of persons taking the test would be acceptable to that university?

Solution

A. Given

*Î¼*= 450,*Ïƒ*= 100. Let*x*be the test score, thenThe area under normal curve between
So, the percentage of the people taking the test score between 400 and 500 is 38.30%.
B. Given
The area under normal curve between
The probability that people taking the test score better is given by:
The probability that people taking the test score worse is given by
So, 3.59 percent people score worse.

C. Given
So,
The people who score more than 480 and acceptable to the university are 61.79%.

*z*= 0 and*z*= 0.5 is 0.1915. Thus, the required probability that the score falls between 400 and 500 is*P*(400 â‰¤*x*â‰¤ 500) =*P*(-0.5 â‰¤*z*â‰¤ 0.5) = 0.1915 + 0.1915 = 0. 3830.*x*= 630,*Î¼*= 450,*Ïƒ*= 100. Let*x*be the test score, then*z*= 0 and*z*= 1.8 is 0.4641.*P*(*x*â‰¥ 630) =*P*(*z*â‰¥ 1.8) = 0.5000 + 0.4641 = 0.9640,*i.e.*, 96.40% people score better.*P*(*x*â‰¥ 630) =*P*(*z*â‰¥ 1.8) = 0.5000 - 0.4641 = 0.0.0359.C. Given

*x*= 480,*Î¼*= 450,*Ïƒ*= 100. Let*x*be the test score. Then, The area under normal curve between*z*= 0 and*z*= 0.30 is 0.1179.*P*(*x*â‰¥ 480) =*P*(*z*â‰¥ 0.30) = 0.5000 + 0.1179 = 0.6179.

Example

The customer accounts of a certain departmental store have an average balance of â‚¹ 480 and a standard deviation of â‚¹160. Assuming that the balance is normally distributed,

a. What is the proportion of the accounts is over â‚¹600?

b. What is the proportion of the accounts is between â‚¹240 and 360?

a. What is the proportion of the accounts is over â‚¹600?

b. What is the proportion of the accounts is between â‚¹240 and 360?

Solution

Let the random variable
a. Hence, 22.66% of the accounts have a balance in excess of â‚¹600.
b. Probability that the accounts lie between â‚¹400 and â‚¹600 is given by

*x*be the valance of the customer accounts. Variable x is normally distributed with*Î¼*= 480 and*Ïƒ*= 160. The standard normal variable is*P*(400 â‰¤*x*â‰¤ 600).Then,
Hence, 46.69% of the accounts have an average balance between â‚¹400 and â‚¹600.

*P*(400 <*x*< 600) =*P*(-0.5 <*x*< 0.75) = 0.1915 + 0.2734 = 0.4649

Example

An aptitude test for selecting an officer in a bank was conducted on 1000 candidates. The average score was 42 and the standard deviations of the scores is 24. Assuming the normal distribution of the scores, find

a. The no. of candidates whose scores exceed 58.

b. The no. of candidates whose scores lie between 30 and 56.

a. The no. of candidates whose scores exceed 58.

b. The no. of candidates whose scores lie between 30 and 56.

Solution

A. Number of candidates whose scores exceed 58:

Area to the right of z = 0.667 under the normal curve is (0.5 â€“ 0.2476) = 0.2524.

Number of candidates whose scores exceed 60 is 100 x 0.2524 = 252.4 or 252.

b. Number of candidates whose scores lie between 30 and 56. Standard normal variate corresponding to 30 is

Standard normal variate corresponding to 66,

Area between z = -0.5 and z = 1 is 0.1915 + 0.3413 = 0.5328.

Number of candidates whose scores lie between 30 and 66 is

1000 x 0.5328 = 532.8 or 533.

Area to the right of z = 0.667 under the normal curve is (0.5 â€“ 0.2476) = 0.2524.

Number of candidates whose scores exceed 60 is 100 x 0.2524 = 252.4 or 252.

b. Number of candidates whose scores lie between 30 and 56. Standard normal variate corresponding to 30 is

Standard normal variate corresponding to 66,

Area between z = -0.5 and z = 1 is 0.1915 + 0.3413 = 0.5328.

Number of candidates whose scores lie between 30 and 66 is

1000 x 0.5328 = 532.8 or 533.

Example

The annual commissions per salesperson employed by a pharmaceutical company, which manufactured cough syrup, averaged â‚¹40,000 with a standard deviation of â‚¹5000. What percent of the salespersons earn between â‚¹32,000 and â‚¹42,000?

Solution

Given
= 0.1554 + 0.4452 = 0.6006,

*Î¼*= 40,*Ïƒ*= 5. Thus,*P*(3200 â‰¤*x*â‰¤ 42000) = =*P*[0.40 â‰¤*x*â‰¤ -1.60]*i.e.*, 60% (approx.)