Normal Distribution.pptx

Research Scholar à ABV- Indian Institute of Information Technology and Management , Gwalior India
12 Apr 2023
1 sur 41

Normal Distribution.pptx

• 1. The Normal distribution MBMG-7104/ ITHS-2202/ IMAS-3101/ IMHS-3101 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 2. What is normal distribution? The Normal Distribution, also called the Gaussian Distribution, is the most significant continuous probability distribution. A normal distribution is a  symmetric, bell-shaped curve  that describes the distribution of continuous random variables.  The normal curve describes how data are distributed in a population. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 3. What is normal distribution? ……A normal distribution is a  A large number of random variables are either nearly or exactly represented by the normal distribution  The normal distribution can be used to represent a wide range of data, such as test scores, height measurements, and weights of people in a population. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 4. What is normal distribution? The normal distribution has two parameters,  mean (μ ) and standard deviation (σ) .  It is important to know these two parameters because they are used to calculate probabilities associated with the normal distribution.  Mean : - It is the measure of central tendency, i.e. it provides us an idea of the concentration of the observations about the central part of the distribution.  Standard deviation :- Standard deviation describes the dispersion or spread the variables about the central value. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 5. Normal distribution is defined by its mean and standard dev. E(X)= = Var(X)=2 = Standard Deviation(X)= dx e x x        2 ) ( 2 1 2 1     2 ) ( 2 1 2 ) 2 1 ( 2              dx e x x @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 6. Normal Distribution Definition  The Normal Distribution is defined by the probability density function for a continuous random variable in a system.  Let us say, f(x) is the probability density function and X is the random variable.  Hence, it defines a function which is integrated between the range or interval (x to x + dx), giving the probability of random variable X, by considering the values between x and x+dx. f(x) ≥ 0 ∀ x ϵ (−∞,+∞) And -∞∫+∞ f(x) = 1 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 7. Normal Distribution Formula  The probability density function of normal or gaussian distribution is given by; Where, x = the variable μ = the population mean σ = standard deviation of the population e = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.1415 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 8. Normal Distribution Curve  The random variables following the normal distribution are those whose values can find any unknown value in a given range.  For example, finding the height of the students in the school. Here, the distribution can consider any value, but it will be bounded in the range say, 0 to 6ft. This limitation is forced physically in our query. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 9. Normal Distribution Curve  Whereas, the normal distribution doesn’t even bother about the range. The range can also extend to –∞ to + ∞ and still we can find a smooth curve.  These random variables are called Continuous Variables, and the Normal Distribution then provides here probability of the value lying in a particular range for a given experiment. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 10. Normal Distribution Standard Deviation  Generally, the normal distribution has any positive standard deviation.  The standard deviations are used to subdivide the area under the normal curve. Each subdivided section defines the percentage of data, which falls into the specific region of a graph. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 11. The Standardized Normal Distribution  The standard normal distribution, also called the z-distribution,  is a special normal distribution where the mean is 0 and the standard deviation is 1.  Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z)  To compute normal probabilities need to transform X units into Z units @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 12. The Standardized Normal Distribution  The standard normal distribution, also called the z-distribution,  is a special normal distribution where the mean is 0 and the standard deviation is 1.  Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z)  To compute normal probabilities need to transform X units into Z units @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 13. The Standardized Normal Distribution  Translate from X to the standardized normal (the “Z” distribution) by subtracting the mean of X and dividing by its standard deviation:  Z scores tell you how many standard deviations from the mean each value lies.  Converting a normal distribution into a z-distribution allows you to calculate the probability of certain values occurring and to compare different data sets. The Z distribution always has mean = 0 and standard deviation = 1 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 14. The Standardized Normal Probability Density Function  The formula for the standardized normal probability density function is Wheree = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.14159 Z = any value of the standardized normal distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 15. Finding Normal Probabilities Probability is measured by the area under the curve a b X f(X) P a X b ( ) ≤ ≤ P a X b ( ) < < = (Note that the probability of any individual value is zero) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 16. Probability as Area Under the Curve Probability is measured by the area under the curve The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below f(X) X μ 0.5 0.5 1.0 ) X P(      0.5 ) X P(μ     0.5 μ) X P(     @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 17. Comparing X and Z units Z ₹100 2.0 0 ₹200 ₹X (μ = ₹100, σ = ₹50) (μ = 0, σ = 1) Note that the shape of the distribution is the same, only the scale has changed. We can express the problem in the original units (X in Rs. ) or in standardized units (Z) Example : If X is distributed normally with mean of ₹100 and standard deviation of ₹50, the Z value for X = ₹ 200 is This says that X = ₹200 is two standard deviations (2 increments of ₹50 units) above the mean of ₹100. 𝑍 = 𝑋 − 𝜇 𝜎 = ₹200 − ₹100 ₹50 = 2.0 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 18. General Procedure for Finding Normal Probabilities To find P(a < X < b) when X is distributed normally:  Draw the normal curve for the problem in terms of X  Translate X-values to Z-values  Use the Standardized Normal Table @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 19. The value within the table gives the probability from Z =   up to the desired Z value .9772 2.0 P(Z < 2.00) = 0.9772 The row shows the value of Z to the first decimal point The column gives the value of Z to the second decimal point 2.0 . . . Z 0.00 0.01 0.02 … 0.0 0.1 The Cumulative Standardized Normal table gives the probability less than a desired value of Z (i.e., from negative infinity to Z) The Standardized Normal Table Z 0 2.00 0.9772 Example: P(Z < 2.00) = 0.9772 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 20. The Standardized Normal Table What is the area to the left of Z=1.51 in a standard normal curve? Z=1.51 Z=1.51 Area is 93.45% @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 21. Finding Normal Probabilities  Let X represent the time it takes (in seconds) to download an image file from the internet.  Suppose X is normal distribution with a mean of18.0 seconds and a standard deviation of 5.0 seconds. Find P(X < 18.6) 18.6 X 18.0 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 22. Finding Normal Probabilities Let X represent the time it takes, in seconds to download an image file from the internet. Suppose X is normal with a mean of 18.0 seconds and a standard deviation of 5.0 seconds. Find P(X < 18.6) Z 0.12 0 X 18.6 18 μ = 18 σ = 5 μ = 0 σ = 1 (continued) 0.12 5.0 8.0 1 18.6 σ μ X Z      P(X < 18.6) P(Z < 0.12) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 23. Z 0.12 0.5478 Standardized Normal Probability Table (Portion) 0.00 = P(Z < 0.12) P(X < 18.6) Z .00 .01 0.0 .5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .02 0.1 .5478 Solution: Finding P(Z < 0.12) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 24. Finding Normal Upper Tail Probabilities  Suppose X is normal with mean 18.0 and standard deviation 5.0.  Now Find P(X > 18.6) X 18.6 18.0 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 25. Now Find P(X > 18.6)… Z 0.12 0 Z 0.12 0.5478 0 1.000 1.0 - 0.5478 = 0.4522 P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12) = 1.0 - 0.5478 = 0.4522 Finding Normal Upper Tail Probabilities … @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 26. Finding a Normal Probability Between Two Values  Suppose X is normal with mean 18.0 and standard deviation 5.0. Find P(18 < X < 18.6) P(18 < X < 18.6) = P(0 < Z < 0.12) Z 0.12 0 X 18.6 18 0 5 8 1 18 σ μ X Z      0.12 5 8 1 18.6 σ μ X Z      Calculate Z-values: @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 27. Z 0.12 0.0478 0.00 = P(0 < Z < 0.12) P(18 < X < 18.6) = P(Z < 0.12) – P(Z ≤ 0) = 0.5478 - 0.5000 = 0.0478 0.5000 Z .00 .01 0.0 .5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .02 0.1 .5478 Standardized Normal Probability Table (Portion) Solution: Finding P(0 < Z < 0.12) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 28. Probabilities in the Lower Tail  Suppose X is normal with mean 18.0 and standard deviation 5.0.  Now Find P(17.4 < X < 18) X 17.4 18.0 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 29. Probabilities in the Lower Tail Now Find P(17.4 < X < 18)… X 17.4 18.0 P(17.4 < X < 18) = P(-0.12 < Z < 0) = P(Z < 0) – P(Z ≤ -0.12) = 0.5000 - 0.4522 = 0.0478 (continued) 0.0478 0.4522 Z -0.12 0 The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 30. Empirical Rule μ ± 1σ encloses about 68.26% of X’s f(X) X μ μ+1σ μ-1σ What can we say about the distribution of values around the mean? For any normal distribution: σ σ 68.26% @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 31. The Empirical Rule  μ ± 2σ covers about 95.44% of X’s  μ ± 3σ covers about 99.73% of X’s x μ 2σ 2σ x μ 3σ 3σ 95.44% 99.73% (continued) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 32. Given a Normal Probability Find the X Value  Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula: Zσ μ X   @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 33. Finding the X value for a Known Probability Example:  Let X represent the time it takes (in seconds) to download an image file from the internet.  Suppose X is normal with mean 18.0 and standard deviation 5.0  Find X such that 20% of download times are less than X. X ? 18.0 0.2000 Z ? 0 (continued) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 34. Find the Z value for 20% in the Lower Tail  20% area in the lower tail is consistent with a Z value of -0.84 Z .03 -0.9 .1762 .1736 .2033 -0.7 .2327 .2296 .04 -0.8 .2005 Standardized Normal Probability Table (Portion) .05 .1711 .1977 .2266 … … … … X ? 18.0 0.2000 Z -0.84 0 1. Find the Z value for the known probability @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 35. Finding the X value 2. Convert to X units using the formula: 8 . 13 0 . 5 ) 84 . 0 ( 0 . 18 Zσ μ X       So 20% of the values from a distribution with mean 18.0 and standard deviation 5.0 are less than 13.80 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 36. Exercise Q.1 The mean of the weight of the students in this class is 80 kg, the standard deviation (SD) is 10.5 kg, find the probability for :- • Find P(X < 85.5) • Find P(X > 85.5) • Find P(80 < X < 85.5) • Find P(75 < X < 80) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 37. Exercise Q.2 IQ tests are measured on a scale which is N (100, 15). A woman wants to form an 'Eggheads Society' which only admits people with the top 1% of IQ scores. What would she have to set as the cut-off point in the test to allow this to happen? @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 38. Exercise Q3. A manufacturer does not know the mean and SD of the diameters of ball bearings he is producing. However, a sieving system rejects all bearings larger than 2.4 cm and those under 1.8 cm in diameter. Out of 1000 ball bearings 8% are rejected as too small and 5.5% as too big. What is the mean and standard deviation of the ball bearings produced? @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 39. Exercise Solution 3 : Assume a normal distribution of N( 1 − 0.08) So 1-0.08 = 0.92 = 92% area in the lower tail is consistent with a Z value of = Z value for 0.92 = 1.4 so 1.8 is 1.4 standard deviations below mean Similarly for N ( 1 − 0.055) , area in the upper tail is 1-0.055 = 0.945, so from z-table the value of z = 1.6 , so 2.4 is 1.6 standard deviations above the mean. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 40. Exercise @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM