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### Normal probability distribution

• 1. NADEEM UDDIN ASSOCIATE PROFESSOR OF STATISTICS NORMAL PROBABILITY DISTRIBUTION
• 2. A continuous random variable is a random variable where the data can take infinite values. For example measuring the time taken for something to be done is continuous random variable. A continuous random variable can be describe by a graph and by a function. A normal distribution, sometimes called the bell curve, The bell curve is symmetrical. Half of the data will fall to the left of the mean and half will fall to the right.
• 3. Normal Distribution: The normal distribution refers to a family of continuous probability distributions described by the normal equation. The normal distribution is defined by the following equation.  2 2 2 1 f(x)= , 2 x e x           
• 4. Standard Normal Distribution: The standard normal distribution is a special case of the normal distribution. It is the distribution that occurs when a normal random variable has a mean of zero and a standard deviation of one. The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation: . . x S N V Z      Where X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X.
• 6. Using Standard Normal Variable Example-1 Given a normal distribution with µ = 40 and σ = 6 , find (a). the area below 32, (b). the area above 27, (c). the area between 42 and 51.
• 7. Solution : Since µ = 40 and σ = 6, X ̴ N(40,62) You need to find p(x < 32).To be able to use the standard normal table, standardize the x variable by subtracting the mean 40 then dividing by the standard deviation, 6.Appling this to both sides of the inequality x<32. (a). 𝑃 𝑥 < 32 = 𝑃 𝑥−𝜇 σ < 32−𝜇 σ = 𝑃 𝑍 < 32 − 40 6 = 𝑃 𝑍 < −1.33 = 0.0918 32 40
• 8. (b). 𝑃(𝑥 > 27) = 𝑃 𝑥−𝜇 σ > 27−𝜇 σ = 𝑃 𝑍 > 27−40 6 = 𝑃 𝑍 > −2.17 =1 - 𝑃 𝑍 < −2.17 = 1 − 0.0150 =0.9850 27 40
• 9. (c). 𝑃(42 < 𝑥 < 51) = 𝑃 42 − 𝜇 σ < 𝑥 − 𝜇 σ < 51 − 𝜇 σ = 𝑃 42−40 6 < 𝑍 < 51−40 6 = 𝑃 0.33 < 𝑍 < 1.83 = 𝑃 0 < 𝑍 < 1.83 − 𝑃 0 < 𝑍 < 0.33 = 0.4664 − 0.1293 = 0.3371 40 42 51
• 10. Example-2 The life of a certain make of saver bulb is known to be normally distributed with a mean life of 2000 hours and a standard deviation of 120 hours. In a batch of 5000 saver bulbs, estimate the number of saver bulbs that the life of such a bulb will be (a)Less than 1950 hours’ (b)Greater than 2150 hours’ (c)From 1850 hours to 2090 hours.
• 11. 1950 2000 Solution : (a). 𝑃(𝑥 < 1950) = 𝑃 𝑥−𝜇 σ < 1950−𝜇 σ = 𝑃 𝑍 < 1950−2000 120 = 𝑃 𝑍 < −0.42 = 0.3372 Number of saver bulb life less than 1950 hours is ; 0.3372×5000 = 1686 bulbs.
• 12. 2000 2150 (b). 𝑃(𝑥 > 2150) = 𝑃 𝑥−𝜇 σ > 2150−𝜇 σ = 𝑃 𝑍 > 2150−2000 120 = 𝑃 𝑍 > 1.25 =1 - 𝑃 𝑍 < 1.25 = 1 − 0.8944 = 0.1056 Number of saver bulb life greater than 2150 hours is ; 0.1056×5000 = 528 bulbs.
• 13. 1850 2000 2090 (c). 𝑃(1850 < 𝑥 < 2090) = 𝑃 1850 − 𝜇 σ < 𝑥 − 𝜇 σ < 2090 − 𝜇 σ = 𝑃 1850−2000 120 < 𝑍 < 2090−2000 120 = 𝑃 −1.25 < 𝑍 < 0.75 = 𝑃 0 < 𝑍 < 0.75 − 𝑃 −1.25 < 𝑍 < 0 = 0.7734 − 0.1056 = 0.6678 Number of saver bulb life between 1850 hours to 2090 hours is 0.6678×5000 = 3339 bulbs.
• 14. Example-3 If X is normally distributed with parameters µ = 6 and 𝜎2 = 16 find (a) 𝑃 𝑥 − 4 ˂ 8 (b) 𝑃( 𝑥 − 4 > 8) (c) 𝑃( 𝑥 − 4 > 8) by using complementary law. Solution: (a) 𝑃 𝑥 − 4 < 8 = 𝑃(−4 ˂ 𝑥˂12) 𝑃 𝑥 − 4 < 8 = 𝑃( −4−𝜇 𝜎 ˂ 𝑥−𝜇 𝜎 ˂ 12−𝜇 𝜎 ) 𝑃 𝑥 − 4 < 8 = 𝑃( −4−6 4 ˂ 𝑍 ˂ 12−6 4 ) 𝑃 𝑥 − 4 < 8 = 𝑃(−2.50˂ 𝑍 ˂1.50)
• 15. 𝑃 𝑥 − 4 < 8 = 𝑃 𝑍 ˂ 1.50 − 𝑃 𝑍 ˂ − 2.50 𝑃 𝑥 − 4 < 8 = 0.9332 − 0.0062 𝑃 𝑥 − 4 < 8 = 0.9270 (b) 𝑃 𝑥 − 4 > 8 = 1 − 𝑃 𝑥 − 4 < 8 𝑃 𝑥 − 4 > 8 = 1 − 𝑃(−4 ˂ 𝑥˂12) 𝑃 𝑥 − 4 > 8 = 1 − 𝑃( −4−𝜇 𝜎 ˂ 𝑥−𝜇 𝜎 ˂ 12−𝜇 𝜎 ) 𝑃 𝑥 − 4 > 8 = 1 − 𝑃( −4−6 4 ˂ 𝑍 ˂ 12−6 4 ) 𝑃 𝑥 − 4 > 8 = 1 − 𝑃(−2.50˂ 𝑍 ˂1.50)
• 16. 𝑃 𝑥 − 4 > 8 = 1 − {𝑃 𝑍 ˂ 1.50 − 𝑃 𝑍 ˂ − 2.50 } 𝑃 𝑥 − 4 > 8 = 1 − (0.9332 − 0.0062) 𝑃 𝑥 − 4 > 8 = 1 − 0.9270 𝑃 𝑥 − 4 > 8 = 0.0730 (c) 𝑃 𝑥 − 4 > 8 = 1 − 𝑃 𝑥 − 4 < 8 𝑃 𝑥 − 4 > 8 = 1 − 𝑃 −4 ˂ 𝑥˂12 𝑃 𝑥 − 4 > 8 = 1 − 0.9270 (from part a) 𝑃 𝑥 − 4 > 8 = 0.0730
• 17. Use of Normal Approximation to The Binomial Distribution We can use the normal distribution as a close approximation to the binomial distribution when n is large and p is moderate (not to far from 0.5) . OR We can use the normal distribution as a close approximation to the binomial distribution whenever n×p ≥ 5 and n×q ≥ 5.
• 18. As we know that the binomial distribution is a discrete distribution and the normal distribution is a continuous distribution, When we use normal approximation we should use the following continuity correction rules. (i) P(x1 ≤ X ≤ x2) = P(x1- 0.5 < X < x2+ 0.5) (ii) P(x1 < X < x2) = P(x1+ 0.5 < X < x2- 0.5) (iii) p(X > x) = p(X > x+0.5) (iv) p(X ≥ x) = p(X > x – 0.5) (v) p(X < x) = p(X < x – 0.5 ) (vi) p(X ≤ x) = p(X < x+0.5) (vii) p(X = x) = P(x1- 0.5 < X < x2+ 0.5)
• 19. Example-4 In a sack of mixed grass seeds, the probability that a seed is ryegrass is 0.35, find the probability that in a random sample of 400 seeds from this sack, (a). less than 120 are ryegrass seeds, (b). more than 160 are ryegrass seeds, (c). between 120 and 150 are ryegrass (d). 150 are ryegrass. Solution: n = 400 ; p = 0.35 µ = np = 400 × 0.35 = 140 σ = 𝑛𝑝𝑞 = 400 × 0.35 × 0.65 = 9.54
• 20. (a). P 𝒙 < 𝟏𝟐𝟎 = 𝑷 𝒙 < 𝟏𝟏𝟗. 𝟓 = 𝑷 𝒙−𝝁 σ < 𝟏𝟏𝟗.𝟓−𝝁 σ = 𝑷 𝒁 < 𝟏𝟏𝟗.𝟓−𝟏𝟒𝟎 𝟗.𝟓𝟒 = 𝑷 𝒁 < −𝟐. 𝟏𝟒𝟗 = 𝟎. 𝟎𝟏𝟓𝟖 (b). P 𝒙 > 𝟏𝟔𝟎 = 𝑷 𝒙 > 𝟏𝟔𝟎. 𝟓 = 𝑷 𝒙−𝝁 σ > 𝟏𝟔𝟎.𝟓−𝝁 σ = 𝑷 𝒁 > 𝟏𝟔𝟎.𝟓−𝟏𝟒𝟎 𝟗.𝟓𝟒 = 𝑷 𝒁 > 𝟐. 𝟏𝟓 = 𝟏 − 𝑷 𝒁 < 𝟐. 𝟏𝟓 = 𝟏 − 𝟎. 𝟗𝟖𝟒𝟐 = 𝟎. 𝟎𝟏𝟓𝟖 120 140 140 160
• 21. (c). P 120 < 𝑥 < 150 = 𝑃 120.5 < 𝑥 < 149.5 = 𝑃 120.5−140 9.54 < 𝑍 < 149.5−140 9.54 = 𝑃 −2.04 < 𝑍 < 0.99 = 𝑃 0 < 𝑍 < 0.99 − 𝑃 −2.04 < 𝑍 < 0 = 0.8389 − 0.0207 = 0.8182 120 140 150
• 22. (d) P x = 150 = P 149.5 < x < 150.5 P x = 150 = P 149.5−μ σ < x−μ σ < 150.5−μ σ P x = 150 = P 149.5−140 9.54 < Z < 150.5−140 9.54 P x = 150 = P 0.99 < Z < 1.10 P x = 150 = P Z < 1.10 − P Z < 0.99 P x = 150 = 0.8643 − 0.8389 P x = 150 = 0.0254 140 149.5 150.5
• 23. Use of Normal Approximation to The Poisson Distribution We can use the normal distribution as a close approximation to the Poisson distribution whenever Mean is large (more than 15). As we know that the Poisson distribution is a discrete distribution and the normal distribution is a continuous distribution, When we use normal approximation we should use the following continuity correction rules. (i) P(x1 ≤ X ≤ x2) = P(x1- 0.5 < X < x2+ 0.5) (ii) P(x1 < X < x2) = P(x1+ 0.5 < X < x2- 0.5) (iii) p(X > x) = p(X > x+0.5) (iv) p(X ≥ x) = p(X > x – 0.5) (v) p(X < x) = p(X < x – 0.5 ) (vi) p(X ≤ x) = p(X < x+0.5) (vii) p(X = x) = P(x1- 0.5 < X < x2+ 0.5)
• 24. Example-5 In a certain city area the number of accidents occurring in a month follows a Poisson distribution with mean 3. Find the following probability that there will be accidents during 10 months. (a) at least 35 accidents (b) Less 30 accidents (c) Between 25 and 35 (d) 35 accidents. Solution: µ = 3×10 = 30 (in 10 months) As we know in Poisson distribution mean is equal to variance Therefore 𝛔 𝟐 = 𝟑𝟎 σ = 5.48 (a) 𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐱 > 𝟑𝟒. 𝟓 𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐱−𝛍 𝛔 > 𝟑𝟒.𝟓−𝛍 𝛔
• 25. 𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐙 > 𝟑𝟒. 𝟓 − 𝟑𝟎 𝟓. 𝟒𝟖 𝐏 𝐱 ≥ 𝟑𝟓 = 𝐏 𝐙 > 𝟎. 𝟖𝟐 𝐏 𝐱 ≥ 𝟑𝟓 = 𝟏 − 𝐏 𝐙 < 𝟎. 𝟖𝟐 𝐏 𝐱 ≥ 𝟑𝟓 = 𝟏 − 𝟎. 𝟕𝟗𝟑𝟗 𝐏 𝐱 ≥ 𝟑𝟓 = 𝟎. 𝟐𝟎𝟔𝟏 (b) 𝐏 𝐱 < 𝟐𝟓 = (𝐱 < 𝟐𝟒. 𝟓) 𝐏 𝐱 < 𝟐𝟓 = ( 𝐱−𝛍 𝛔 < 𝟐𝟒.𝟓−𝛍 𝛔 ) 𝐏 𝐱 < 𝟐𝟓 = (𝐙 < 𝟐𝟒.𝟓−𝟑𝟎 𝟓.𝟒𝟖 ) 𝐏 𝐱 < 𝟐𝟓 = (𝐙 < −𝟏. 𝟎) 𝐏 𝐱 < 𝟐𝟓 = 𝟎. 𝟏𝟓𝟖𝟕
• 26. (c) 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 𝟐𝟓. 𝟓 < 𝐱 < 𝟑𝟒. 𝟓 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 𝟐𝟓.𝟓−𝝁 𝝈 < 𝒙−𝝁 𝝈 < 𝟑𝟒.𝟓−𝝁 𝝈 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 𝟐𝟓.𝟓−𝟑𝟎 𝟓.𝟒𝟖 < 𝐙 < 𝟑𝟒.𝟓−𝟑𝟎 𝟓.𝟒𝟖 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 −𝟎. 𝟖𝟐 < 𝐙 < 𝟎. 𝟖𝟐 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝐏 𝒁 < 𝟎. 𝟖𝟐 − 𝐏 𝒁 < −𝟎. 𝟖𝟐 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝟎. 𝟕𝟗𝟑𝟗 − 𝟎. 𝟐𝟎𝟔𝟏 𝐏 𝟐𝟓 < 𝐱 < 𝟑𝟓 = 𝟎. 𝟓𝟖𝟕𝟖
• 27. (d) P x = 35 = P 34.5 < x < 35.5 P x = 35 = P 34.5−μ σ < x−μ σ < 35.5−μ σ P x = 35 = P 34.5−30 5.48 < Z < 35.5−30 5.48 P x = 35 = P 0.82 < Z < 1.00 P x = 35 = P Z < 1.00 − P Z < 0.82 P x = 35 = 0.8413 − 0.7939 P x = 35 = 0.0474 30 34.5 35.5
• 28. Using the Table in Reverse for any normal variable x Example-6 The time taken by the milkman to deliver to the high street is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. He delivers milk every day. Estimate the time of day during the year. (a) Longer than 0.0062 . (b) Less than 0.1587.
• 29. Solution: X is the time to delivers milk everyday. X ̴ N(12,22) (a) Given p(x > k) = 0.0062 Standardising 𝑷 𝒙−𝝁 σ > 𝒌−𝝁 σ = 𝟎. 𝟎𝟎𝟔𝟐 𝑷 𝒛 > 𝒌−𝟏𝟐 𝟐 = 𝟎. 𝟎𝟎𝟔𝟐 Since 𝒌−𝟏𝟐 𝟐 = 𝟐. 𝟓𝟎 (The value of z at 0.0062 from normal table is 2.50) k= 2.50×2+12 = 17 minute Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 -2.6 .00466 .00453 .00440 .00427 .00415 .00402 .00391 .00379 .00368 .00357 -2.5 .00621 .00604 .00587 .00570 .00554 .00539 .00523 .00508 .00494 .00480 -2.4 .00820 .00798 .00776 .00755 .00734 .00714 .00695 .00676 .00657 .00639
• 30. (b) Given p(x < k) = 0.1587 Standardrising 𝑷 𝒙−𝝁 σ < 𝒌−𝝁 σ = 𝟎. 𝟏𝟓𝟖𝟕 𝑷 𝒛 < 𝒌−𝟏𝟐 𝟐 = 𝟎. 𝟏𝟓𝟖𝟕 Since 𝒌−𝟏𝟐 𝟐 = −𝟏. 𝟎 (The value of z at 0.1587 from normal table is -1.0) k= (-1.0)×2+12 = 10 minute Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 -3.9 .00005 .00005 .00004 .00004 .00004 .00004 .00004 .00004 .00003 .00003 -1.0 .15866 .15625 .15386 .15151 .14917 .14686 .14457 .14231 .14007 .13786 -0.9 .18406 .18141 .17879 .17619 .17361 .17106 .16853 .16602 .16354 .16109
• 31. Example-7 The length of certain items follow a normal distribution with mean  cm and standard deviation 6cm, it is known that 95% of the items have length less than 82cm. Find the value of the mean . Solution: P(x  82) = 0.95 P x−μ σ < k−μ σ = 0.95 P Z < 82−μ 6 = 0.95 82−μ 6 = 1.65 82 - µ =(1.65)×6 µ = 82 – 9.9 = 72
• 32. Example-8 x~N(100, σ2 ) and P(x ˂ 106) = 0.8849. find the standard deviation σ. Solution: P(x  106) = 0.8849 P x − μ σ < 106 − 100 σ = 0.8849 P Z < 6 σ = 0.8849 6 σ = 1.20 6 1.20 = σ σ = 5
• 33. Example-9 In a normal distribution, 5% of the items are up to 45 and 75% are below 65. (a)Find mean and standard deviation, (b)Find P(45 ˂ x ˂ 65). Solution: (a) P 𝑥 > 45 = 0.05 P 𝑥−𝜇 𝜎 > 45−𝜇 𝜎 = 0.05 P 𝑍 > 45−𝜇 𝜎 = 0.05 45−𝜇 𝜎 = −1.65 (from Z- table) 45 - µ = -1.65 𝜎 ------(i)
• 34. 𝑷 𝒙 < 𝟔𝟓 = 𝟎. 𝟕𝟓 P 𝒙−𝝁 𝝈 < 𝟔𝟓−𝝁 𝝈 = 𝟎. 𝟕𝟓 P 𝒁 < 𝟔𝟓−𝝁 𝝈 = 𝟎. 𝟕𝟓 𝟔𝟓−𝝁 𝝈 = 𝟎. 𝟔𝟖 (from Z- table) 65 - µ = 0.68 𝝈 ------(ii) Subtract (i) from (ii) 65 - µ = 0.68 𝝈 45 - µ = -1.65 𝝈 20 = 2.33 𝝈 𝝈 = 𝟐𝟎 𝟐. 𝟑𝟑 = 𝟖. 𝟓𝟖
• 35. Put 𝜎 = 8.58 in (ii) 65 - µ = 0.68 (8.58) 65 - µ = 5.83 µ = 59.17 (b) P 45 < 𝑥 < 65 = 𝑷 𝟒𝟓−𝝁 𝝈 < 𝒙−𝝁 𝝈 < 𝟔𝟓−𝝁 𝝈 = 𝑷 𝟒𝟓−𝟓𝟗.𝟏𝟕 𝟖.𝟓𝟖 < 𝒁 < 𝟔𝟓−𝟓𝟗.𝟏𝟕 𝟖.𝟓𝟖 = 𝑷 −𝟏. 𝟔𝟓 < 𝒁 < 𝟎. 𝟔𝟖 = 𝑷 𝒁 < 𝟎. 𝟔𝟖 − 𝑷 𝒁 < −𝟏. 𝟔𝟓 = 0.7517 – 0.0495 = 0.7022
• 36. Example-10 In a normal distribution with µ = 100 and σ = 15 find (a) 𝑄1 𝑎𝑛𝑑𝑄3 (b) 𝐷7 (c) 𝑃95 Solution: (a) 𝑄1 = First quartile, (the points below which 25% area lies and above which 75% area lies) Thus P(X ≤ 𝑄1) = 0.25 𝑃 𝑥−𝜇 𝜎 ≤ 𝑄1−𝜇 𝜎 = 0.25 𝑃 𝑍 ≤ 𝑄1−100 15 = 0.25 𝑃 𝑍 ≤ −0.67 = 0.25 (from Z- table)
• 37. Hence 𝑄1−100 15 = −0.67 𝑄1 = 100 − (0.67)(15) 𝑄1 = 89.95 and 𝑄3 = Third quartile, (the points below which 75% area lies and above which 25% area lies) Thus P(X ≤ 𝑄3) = 0.75 𝑃 𝑥−𝜇 𝜎 ≤ 𝑄3−𝜇 𝜎 = 0.75 𝑃 𝑍 ≤ 𝑄3−100 15 = 0.75 𝑃 𝑍 ≤ 0.67 = 0.75 (from Z- table)
• 38. Hence 𝑄3−100 15 = 0.67 𝑄3 = 100 + (0.67)(15) 𝑄3 = 110.05 (b) 𝐷7 = 7th deciles, (the points below which 70% area lies and above which 30% area lies) Thus P(X ≤ 𝐷7) = 0.70 𝑃 𝑥−𝜇 𝜎 ≤ 𝐷7−𝜇 𝜎 = 0.70 𝑃 𝑍 ≤ 𝐷7−100 15 = 0.70 𝑃 𝑍 ≤ 0.52 = 0.70 (from Z- table)
• 39. Hence 𝐷7−100 15 = 0.52 𝐷7 = 100 + (0.52)(15) 𝐷7 = 107.8 (c) 𝑃95 = 95th percentile, (the points below which 95% area lies and above which 5% area lies) Thus P(X ≤ 𝑃95) = 0.95 𝑃 𝑥−𝜇 𝜎 ≤ 𝑃95−𝜇 𝜎 = 0.95 𝑃 𝑍 ≤ 𝑃95−100 15 = 0.95 𝑃 𝑍 ≤ 1.65 = 0.95 (from Z- table)
• 40. Hence 𝑃95−100 15 = 1.65 𝑃95 = 100 + 1.65 15 𝑃95 = 124.75
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