CHAPTER 7: NORMAL DISTRIBUTION Mr. Mark Anthony Garcia, M.S. Mathematics Department De La Salle University CHAPTER 7: NORMAL DISTRIBUTION

Normal Distribution A normal distribution is a probability distribution that plots all of its values in a symmetrical fashion and most of the results are situated around the probability's mean. Values are equally likely to plot either above or below the mean. Grouping takes place at values that are close to the mean and then tails off symmetrically away from the mean.

Normal Distribution The normal distribution is represented by a bell-shaped curve called the normal curve and the area under the curve represents the probability of the normal random variable X.

Situation: Normal Distribution Let X be the height of a student from DLSU and X is a normal random variable. Suppose that the mean height of all DLSU students is 𝜇=163𝑐𝑚 with standard deviation 𝜎=20𝑐𝑚. 𝝁−𝝈 143 𝝁+𝝈 183 𝝁−𝟐𝝈 123 𝝁+𝟐𝝈 203 𝝁−𝟑𝝈 103 𝝁+𝟑𝝈 223

Situation: Normal Distribution

Properties of the Normal Curve It is a bell-shaped curve. The mode, which is the point on the horizontal axis where the curve is a maximum, occurs at x = μ. This means that the mean is equal to the mode. The curve is symmetric about a vertical axis through the mean μ. The mean divides the set of data into two equal parts. This means that the mean is equal to the median.

Properties of the Normal Curve The normal curve approaches the horizontal axis asymptotically as we proceed in either direction away from the mean. (The graph approaches the x-axis but the graph will never intersect the x-axis). The total area under the curve and above the horizontal axis is equal to 1.

Formula: Normal Distribution The formula for the normal distribution is given by

Comparing Normal Curves Consider the figure below.

Comparing Normal Curves Observe that the blue, red and yellow normal curves have the same mean because they are centered at 𝜇=0 but with different heights because of different frequencies. However, the green normal curve has mean 𝜇=−2.

Comparing normal curves Moreover, all the normal curves shown have different variances which measures the dispersion or spread of the values in the data set. This is illustrated by the width of the curve. It can be seen from the figure, that the yellow normal curve has the largest width and with variance 𝜎 2 =5.

Comparing Normal Curves To avoid having normal distributions with different means and standard deviations, we convert the normal random variable X into the standard normal random variable Z. The standard normal random variable Z has mean equal to zero (𝜇=0) and standard deviation equal to one (𝜎=1).

Standard Normal Distribution To convert the values of the normal random variable X to the standard normal random variable Z, we use the formula given by 𝑍= 𝑋−𝜇 𝜎 .

Example: From X to Z Suppose that X is the normal random variable with mean 𝜇=550 and standard deviation 𝜎=150. What is the probability that X is greater than 600? In symbols, we have 𝑃(𝑋>600). Using the formula 𝑍= 𝑋−𝜇 𝜎 , substitute 𝑋=600, 𝜇=550 and 𝜎=150.

Example 1: Normal Distribution Hupper Corporation produces many types of softdrinks, including Orange Cola. The filling machines are adjusted to pour 12 ounces of soda into each 12-ounce can of Orange cola. However, the actual amount of soda poured into each can is not exactly 12 ounces; it varies from can to can. It has been observed that the net amount of soda in such a can has a normal distribution with a mean of 12 ounces and a standard deviation of 0.015 ounce.

Example 1: Normal Distribution What is the probability that a randomly selected can of Orange Cola contains 11.97 to 11.99 ounces of soda? Probability for a Range From X Value 11.97 To X Value 11.99 Z Value for 11.97 -2 Z Value for 11.99 -0.666667 P(X<=11.97) 0.0228 P(X<=11.99) 0.2525 P(11.97<=X<=11.99) 0.2297

Example 1: Normal Distribution The probability that a can of Orange Cola will have between 11.97 to 11.99 ounces is 0.2297. If there are 1000 cans of Orange Cola, how many of the 1000 cans will have between 11.97 to 11.99 ounces? The number of cans is 0.2297 1000 =229.7 or approximately 230 cans.

Example 1: Normal Distribution Suppose we delete the last row of the PhStat output. How do we get 𝑃(11.97≤𝑋≤11.99)? Probability for a Range From X Value 11.97 To X Value 11.99 Z Value for 11.97 -2 Z Value for 11.99 -0.666667 P(X<=11.97) 0.0228 P(X<=11.99) 0.2525

Example 1: Normal Distribution To determine the probability, we have 𝑃 𝑋≤11.99 −𝑃 𝑋≤11.97 0.2525−0.0228=0.2297

Example 1: Normal Distribution What percentage of the cola contains at least 12.025 ounces of soda? 𝑃 𝑋>12.025 =0.0478 Probability for X <= X Value 12.025 Z Value 1.6666667 P(X<=12.025) 0.9522096 Probability for X > X Value 12.025 Z Value 1.6666667 P(X>12.025) 0.0478

Example 1: Normal Distribution Suppose that the output below is the only given output. Probability for X <= X Value 12.025 Z Value 1.6666667 P(X<=12.025) 0.9522096

Example 1: Normal Distribution 𝑃 𝑋>12.025 =1−𝑃(𝑋≤12.025) 𝑃 𝑋>12.025 =1−0.9522=0.0478 The probability that the cola will have at least 12.025 ounces of soda is 0.0478.

Example 2: Normal Distribution The price of diesel oil over the past 24 months is normally distributed with a mean of 41 pesos per liter and standard deviation of 5 pesos per liter.

Example 2: Normal Distribution What is the probability that the price of diesel is at most 34 pesos per liter? 𝑃 𝑋≤34 =0.0808 Probability for X <= X Value 34 Z Value -1.4 P(X<=34) 0.0808

Example 2: Normal Distribution For which amount can we find the highest 10% of the diesel prices? Find X and Z Given Cum. Pctage. Cumulative Percentage 10.00% Z Value -1.2816 X Value  34.592 Find X and Z Given Cum. Pctage. Cumulative Percentage 90.00% Z Value 1.2816 X Value 47.408

Example 2: Normal Distribution Since the highest 10% of the diesel prices occurs at the rightmost part of the normal curve, we get the area at the left which is 90%. Thus, we use the table with cumulative percentage equal to 90%.

Example 2: Normal Distribution Getting the X value, we have 47.408. This means that that the probability that diesel prices is more than 47.408 pesos per liter is 10% or 0.10.

Example 2: Normal Distribution Suppose that we delete the last row of the table of cumulative percentage. How do we find the value of X? Find X and Z Given Cum. Pctage. Cumulative Percentage 90.00% Z Value 1.2816 X Value 47.408

Example 2: Normal Distribution Using the formula 𝑍= 𝑋−𝜇 𝜎 , we derive X. Then 𝑋=𝑍𝜎+𝜇. 𝑋= 1.2816 5 +41=47.408

Exercises: Normal Distribution The TV ratings of the show The Big Bang Theory are approximately normally distributed with mean 22.7 and standard deviation 7.4. Probability for X > X Value 26.1 Z Value 0.5 P(X>26.1) 0.3085 Probability for X <= X Value 27.6 Z Value 0.7027027 P(X<=27.6) 0.7588795

Exercises: Normal Distribution What is the probability that for a given day the show The Big Bang Theory will obtain a rating of at least 27.6? Given 15 episodes of the show The Big Bang Theory, how many episodes would get a rating of less than 26.1?

Exercises: Normal Distribution In the November 1990 issue of Chemical Engineering Progress, a study discussed the percent purity of oxygen from a certain supplier. Assume that the mean was 99.61 with a standard deviation of 0.08. Assume that the distribution of percent purity was approximately normal.

Exercises: Normal Distribution What percentage of the purity values would you expect to be between 99.5 and 99.7? Probability for X <= X Value 99.5 Z Value -1.375 P(X<=99.5) 0.0845657 Probability for a Range From X Value 99.5 To X Value 99.7 Z Value for 99.5 -1.375 Z Value for 99.7 1.125 P(X<=99.5) 0.0846 P(X<=99.7) 0.8697 Probability for X > X Value 99.7 Z Value 1.125 P(X>99.7) 0.1303

Exercises: Normal Distribution What purity value would you expect to exceed exactly 5% of the population? Find X and Z Given Cum. Pctage. Cumulative Percentage 5.00% Z Value -1.644854 Find X and Z Given Cum. Pctage. Cumulative Percentage 95.00% Z Value 1.644854