Unformatted text preview: Chapter 8 : Chapter Applications of the Normal Distribution Applications
The normal distribution arises from a very natural, normal situation, such as product quality. Very many other characteristics, such as I.Q; body weight, body height, flight time, monthly recurrent expenses, examination performances, etc, demonstrate normal distributions. Thus, the normal distribution has many applications. Here we shall demonstrate some. Before the examples, let us put down, for the sake of completeness, the mathematical formula for the normal curve (although we shall NOT need it in this course): y= 1 2πσ 2 e −( x−µ )2 2σ 2 ,− < x < ∞ ∞ where μ = centre = mean, σ= standard deviation, standard e = continuous growth base = 2.7183. e = continuous growth base = 2.7183. We write X ~ N(μ,σ²). In application, we are mainly interested in the area A under In application, we are mainly interested in the area A under the curve and lying between μ andμ+zσ: A = P(μ< x <μ+zσ) P( The values of A for various z have been found by The mathematicians and are listed in Table A. The reverse relation, finding z from given Area, is listed in Table B. Table We shall learn how to make use of these two tables. Example 1: Example Circular metal discs are manufactured. The diameter, x, has a normal distribution, with mean μ=15.00 cm, and standard deviation σ=0.10cm. =15.00 =0.10cm. Only discs whose diameters lie between 14.80cm and 15.15cm are regarded as conforming with manufacturing specifications. Find the proportion of items correctly manufactured. Find Solution: The problem says: X ~ N ( 15, 0.1² = 0.01 ) ii.e; X is distributed like a normal model, with its mean .e; equal to 15 and its variance equal to 0.1² = 0.01. equal We want to find P(14.8 < X < 15.15). x2 x1 There are two cutoff values, and , from which we want to find the areas, and respectively. A A1 2 First, we need to find two signal/sigma ratios, and z1 satisfying , z2 Signal = xi − µ Sigma = σ namely , z1 = x1 − µ 14.80 − 15 = = −2 σ 0.1 z2 = x2 − µ σ = 15.15 −15 = 1.5 0.1 For the concept of Signal/Sigma Ratio (i.e; standardized scores) P (14.8 < X < 15.15) = A1 + A2 = P(−2 < Z < 1.5) = 0.4772 + 0.4332, fromTableA = 0.9104 = 91.04%
Note 1: Note 1 To find A1 from 50FH: x? ON MODE C0MP FMLA 05 2 EXE To find A2 from 50FH: FMLA 05 1.5 EXE x?
0.43319 = A2 0.47725 = A1 Note 2: Note If one item is randomly selected from the If (large) batch of manufactured items, then the probability that it is correctly probability manufactured, is 0.9104. Thus the P in the above expression may mean percentage, percentage proportion, or probability, interchangeably. proportion probability
P (14.8 < X <15.15) = A1 + A2 = P ( −2 < Z <1.5) = 0.4772 + 0.4332, fromTableA = 0.9104 = 91.04% Example 2: Example My husband is often late for our appointments. On the whole, he is half an hour late, with a standard deviation of 8 minutes. His arrival time follows a normal distribution. Today, I have invited him to meet me at the lobby of the Mandarin Hotel at 7pm. I am, unfortunately, delayed by traffic, so I will only reach the door of the Mandarin at 7:16pm. What is my feeling at this moment? Why? Solution: Solution Let X = my husband’s arrival time, Let (in minutes), counting from 7pm. Then, μ=30,σ=8. Then, =8. In other words, X ~ N(μ=30,σ² = 8² = 64) In Find P(I arrive later than my husband) = P( X < 16 ) = 0.5 – A1 The cutoff value is The cutoff value is x1 = 16
z1 = x1 − µ 16 30 = = −1.75 σ 8 The corresponding S/S ratio is From Table A, we find P (trouble ) A1 = 0.4599 = (X < ) P 16 X− 30 16 − 30 =( P < ) 8 8 = ( Z < 1.75) P − = .5 − 1 0 A = .5 − .4599, fromTableA 0 0 = .0401 0 I was calm, because the chance that he arrived before me was only 0.0401 (only 4.01% chance). Example 3: Example
Given that X ~N (μ,σ²). Given that X ~N ( Find (a) P(μ-σ< X <μ+σ) (b) P(μ+σ< X <μ+2σ) (c) P(μ+2σ< X <μ+3σ) (d) P(X >μ+3σ) Solution: Solution From Table A, we have these values: From Hence, (a) P ( µ −σ < X < µ +σ ) = P[ ( µ −σ ) − µ = P ( −1 < Z < 1) = 0.3413 + 0.3413 = 0.6826 σ < X −µ σ < ( µ +σ ) − µ σ Alternatively, P( µ − σ < X < µ + σ ) = P( µ − σ < X < µ ) + P( µ < X < µ + σ ) (µ − σ ) − µ µ−µ µ−µ µ +σ = P[ <Z< ] + P[ <Z< ] σ σ σ σ Note : S / S = ±1 = P(−1 < Z < 1) = P(−1 < Z < 0) + P(0 < Z < 1) = 0.3413 + 0.3413 = 0.6826 (b) (b) Solution: Solution P ( µ + σ < X < µ + 2σ ) ( µ + σ ) − µ X − µ ( µ + 2σ ) − µ = P[ < < ] σ σ σ = P (1 < Z < 2) = 0.4772 − 0.3413 = 0.1359 Solution: Solution
(c) (c) P( µ + 2σ < X < µ + 3σ ) ( µ + 2σ ) − µ X − µ ( µ + 3σ ) − µ = P[ < < ] σ σ σ = P(2 < Z < 3) = 0.4987 − 0.4772 = 0.0215 (d) P ( X > µ + 3σ ) X − µ ( µ + 3σ ) − µ = P[ > ] σ σ = P ( Z > 3) = 0.5 − 0.4987 = 0.0013 Example 4: Example Let X ~ N (μ,σ²). Let ²). Find a value “z” such that P(μ – zσ < X <μ + zσ) = 0.95 P(
Solution: Using the normal tables, we need to find P(μ< X <μ+zσ) = 0.475 From Table B, we find z = 1.96, since From 1.96 since P(μ-1.96σ< X <μ+1.96σ) = 0.95 P(
(This value, 1.96, is also obtainable from Table A-backwards!) Solution: Let X = weight of marble (in grams) then X ~ N(μ = ? ,σ² = 0.20²) Find μ such that P( X > 85) = 3% Example 5: Example The weights of ball-bearings (metal marbles) manufactured in a factory have a normal distribution with standard deviation 0.20g. It is desired that only 3% of the marbles weigh more than 85g. At what value(μ) must the manufacturer set At must the quality target? the Now P(X > 85) = 0.03 Now P(X > 85) = 0.03 µ − µ X − µ 85 − µ P[ < < ] = 0.47 Therefore σ σ σ P(μ< X < 85) = 0.47 85 − µ ) = 0 .4 7 Now standardize: P (0 < Z < Now 0 .2 From Table B, the S/S ratio From (z-score) corresponding to an area of 0.47 is z = 1.881 1.881 However, by definition: However, x−µ z= σ 85 − µ 1.881 = 0.2 µ = 85 − 1.881× 0.2 = 84.62 Example 6: Example The length of sardines received by a cannery have a normal distribution with mean 3.44 inches and s.d.=0.23 inches. What percentage of all sardines are shorter than 3.80 inches? than
Solution: Let X = the length of a sardine (in inches) X ~ N(μ = 3.44, σ²=0.23²=0.0529) Find P( X < 3.80) = ? X − µ 3.8 − 3.44 P ( X < 3.8) = P( < ) = P ( Z < 1.5652) = 0.5 + A1 σ 0.23 To find A1, we consult Table A, but we can To find A1, we consult Table A, but we can only obtain: P ( X < 3.8) = P( Z < 1.5652) = 0.5 + ?
By the method of interpolation: By the method of interpolation:
A1 = 0.52 × 0.4418 + 0.48 × 0.4406 = 0.4412 ∴ P( X < 3.8) = P ( Z < 1.5652) = 0.5 + A1 = 0.5 + 0.4412 = 0.9412 = 94.12% Of course, it’s faster to use your calculator! Of course, it’s faster to use your calculator! x? ON MODE FMLA 05 1.5652 EXE
0.44123 = A1 P ( X < 3.8) = P ( Z < 1.5652) = 0.5 + 0.44123 = 0.94123 Simple Exercises Simple
1. The assembly time, X, (in seconds) of a certain production job has a N(27.8,2.52) distribution. Find P(X<24.5) and P(25.0< X< 32.0). 2. The net amount, X (in grams) of instant coffee a filling machine puts into “120g” jars hasa normal distribution with σ=0.2g. =0.2g. If only 3% of the jars are to contain less than 120g of coffee, what must the mean fill be? of 3. Find the area under the standard normal curve 3. Find the area under the standard normal curve that lies: (a) between z=0.38 and z=0.76 (b) between z=2.88 and z=3.12 (c) below z=2.89 (d) above z=1.2345 4. A normal variable, X, has an unknown mean, μ, and unknown standard deviation, σ. It is known that P(X<50)=0.08 and P(100<X)=0.18. Find μand σ. and ...
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