Normal Curve Distribution PDF

NORMAL CURVE DISTRIBUTION (Psychological Statistics – Lecture) Mathematics Learning Area De La Salle Lipa  NORMAL CURVE DISTRIBUTION One of the most important examples of a continuous probability distribution is the normal distribution....

NORMAL CURVE DISTRIBUTION (Psychological Statistics – Lecture) Mathematics Learning Area De La Salle Lipa  NORMAL CURVE DISTRIBUTION One of the most important examples of a continuous probability distribution is the normal distribution. Applications of a normal probability distribution are so numerous that some mathematicians refer to it as “a veritable Boy Scout knife of statistics”. Definition The normal distribution is a probability function that describes how the values of a variable are distributed. It is also known as Gaussian distribution and the bell curve.  NORMAL CURVE DISTRIBUTION The empirical rule tells you what percentage of your data falls within a certain number of standard deviations from the mean: 68% of the data falls within one standard deviation of the mean 95% of the data falls within two standard deviation of the mean 99.7% of the data falls within three standard deviation of the mean  NORMAL CURVE DISTRIBUTION Properties of a Normal Distribution Represented by a bell-shaped curve and its probability distribution is called normal distribution. The mean = median = mode It is symmetrical about the mean. The tails or ends are asymptotic relative to the horizontal axis. The total area under the normal curve is 1 or 100%. The normal curve area maybe subdivided into standard deviations, at least 3 to the left and 3 to the right of the mean.  NORMAL CURVE DISTRIBUTION Standard Normal Distribution A standard normal model is a normal distribution with a mean of zero (0) and a standard deviation of 1.  NORMAL CURVE DISTRIBUTION Standard Score (z-score) To determine the areas under the normal curve, we shall convert original scores into standard score or z-score. This means that the empirical distribution will be standardized to the theoretical normal curve. It is the position of a value of x in terms of the number of standard deviations, which is located from the mean. Formula: Derived Formula: 𝑋−𝜇 (Converting z-score to original (raw) score) 𝑧= 𝜎 𝑋 = 𝑧𝜎 + 𝜇 where: X – Original Score μ – Mean σ – Standard Deviation The normal distribution depends on the standard deviation and mean  NORMAL CURVE DISTRIBUTION Standard Score Example 1: Given the mean (µ) = 85 and standard deviation (σ) = 5, convert the following original scores to z-scores. a. Original Score (X) = 90 b. Original Score (X) = 75 𝑋−𝜇 90−85 𝑋−𝜇 75−85 𝑧= = =1 𝑧= = = −2 𝜎 5 𝜎 5 c. Original Score (X) = 102 d. Original Score (X) = 83 𝑋−𝜇 102−85 𝑋−𝜇 83−85 𝑧= = = 3.4 𝑧= = = −0.4 𝜎 5 𝜎 5 e. Original Score (X) = 86 f. Original Score (X) = 66 𝑋−𝜇 86−85 𝑋−𝜇 66−85 𝑧= = = 0.2 𝑧= = = −3.8 𝜎 5 𝜎 5  NORMAL CURVE DISTRIBUTION Standard Score Example 2: Algebra English a. The table shows the grades of Juan in mean 80 75 Algebra and English. In what subject do SD 5 5 you think Juan performed better? Score 85 82 Solutions: 85−80 z-score of Juan in Algebra = 𝑧Algebra = =1 5 82−75 z-score of Juan in English = 𝑧English = = 1.4 5 Since zEnglish is higher than zAlgebra, we can say that Juan performed better in English test than in Algebra.  NORMAL CURVE DISTRIBUTION Standard Score Example 2: Algebra English b. A classmate of Juan in Algebra has a mean 80 75 z-score of 2.1. What is his test score? SD 5 5 Score 85 82 Solutions: X = 𝑧𝜎 + 𝜇 = 2.1 5 + 80 = 90.5 Juan’s classmate scored 90.5 in their Algebra test.  NORMAL CURVE DISTRIBUTION Getting the areas under the normal curve reminders: 1. The total area under the normal curve is 1 or 100%. 2. Since the normal curve is symmetrical about the mean, then half the normal curve has an area of 0.50. 3. The table you will use gives only the area to the right of the mean. 4. The given area in the table is the area from z = 0 to ± z 5. Area is always + but z can be + or –. NORMAL CURVE DISTRIBUTION  NORMAL CURVE DISTRIBUTION Example 1: Finding the areas under the normal curve is the same as finding the probability. a. Find the area from z = 0 to z = 1.35 Solutions: To use the z-table, start on the left side of the table go down to 1.3 and now at the top of the table, go to 0.05 (this corresponds to the z value of 1.3 + 0.05 = 1.35). The value is 𝑃(0 ≤ 𝑧 ≤ 1.35) = 0.4115  NORMAL CURVE DISTRIBUTION Example 1: Finding the areas under the normal curve is the same as finding the probability. b. Find the area from z = –2 to z = 2.53 Solutions: 𝑃 −2 ≤ 𝑧 ≤ 2.53 = 𝑃 −2 ≤ 𝑧 ≤ 0 + 𝑃(0 ≤ 𝑧 ≤ 2.53) = 0.4772 + 0.4943 𝑃(−2 ≤ 𝑧 ≤ 2.5) = 0.9715  NORMAL CURVE DISTRIBUTION Example 1: Finding the areas under the normal curve is the same as finding the probability. c. Find P(z ≥ –1.73) Solutions: P(z ≥ –1.73) = 𝑃 −1.73 ≤ 𝑧 ≤ 0 + 𝑃(0 ≤ 𝑧 ≤ +∞) = 0.4582 + 0.5000 P(z ≥ –1.73) = 0.9582  NORMAL CURVE DISTRIBUTION Example 1: Finding the areas under the normal curve is the same as finding the probability. d. Find P(0.52 ≤ z ≤ 2.50) Solutions: P(0.52 ≤ z ≤ 2.50) = 𝑃 0 ≤ 𝑧 ≤ 2.50 − 𝑃(0 ≤ 𝑧 ≤ 0.52) = 0.4938 – 0.1985 P(0.52 ≤ z ≤ 2.50) = 0.2853  NORMAL CURVE DISTRIBUTION Example 1: Finding the areas under the normal curve is the same as finding the probability. e. Find P(z ≥ 0.87) Solutions: P(z ≥ 0.87) = 𝑃(0 ≤ 𝑧 ≤ ∞) − 𝑃(0 ≤ 𝑧 ≤ 0.87) = 0.5000 – 0.3078 P(z ≥ 0.87) = 0.1922  NORMAL CURVE DISTRIBUTION Example 2: The IQ of 300 students in a certain high school is approximately normally distributed with μ = 100 and σ = 15. a. What is the probability that a randomly selected student will have an IQ of 115 and above? Solutions: First, determine the z-score of the IQ level. 115 − 100 𝑧= =1 15 So, we are to find P(z ≥ 1). P(z ≥ 1.00) = 𝑃 0 ≤ 𝑧 ≤ ∞ − 𝑃(0 ≤ 𝑧 ≤ 1.00) = 0.5000 – 0.3413 P(z ≥ 0.87) = 0.1587  NORMAL CURVE DISTRIBUTION Example 2: The IQ of 300 students in a certain high school is approximately normally distributed with μ = 100 and σ = 15. b. How many students have an IQ from 85 to 120? Solutions: First, determine the z-scores of the IQ levels. 85−100 120−100 𝑧= = −1 𝑧= = 1.33 15 15 So, we are to find P(–1 ≤ z ≤ 1.33). P(–1≤ z ≤ 1.33) = 𝑃 −1 ≤ 𝑧 ≤ 0 + 𝑃(0 ≤ 𝑧 ≤ 1.33) = 0.3413 + 0.4082 P(–1≤ z ≤1.33) = 0.7495 Number of students = 300 × 0.7495 = 224.85 ~ 225 students  NORMAL CURVE DISTRIBUTION Test Your Understanding 11: From a large number of beauty soap bars, samples were selected at random and produced a mean weight of 90 grams and a standard deviation of 5 grams. Assuming that the weights are normally distributed and that only soap bars weighing between 80 and 97 grams are acceptable, what percent of the soap bars is to be rejected?