Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications F7-MS-Ch16 - 1 16 The normal distribution (正態分佈) § 16.1 The normal distribution 正態分佈 It is a continuous probability distribution (連續概率分佈) with mean and variance 2 (or standard deviation ) such that its p.d.f. f (x) is given by f ( x) 1 2 ( x )2 e 2 2 for all values of x . If x is distributed in this way, we write x ~ N ( , 2 ) § 16.2 Properties of normal distribution 正態分佈的特性 (a) The mean (平均數), median (中位數) and the mode (眾數) are equal. (b) The curve is bell-shaped (鐘形) and symmetric (對稱) about the mean. (c) The curve approaches (趨近) the x-axis (or z-axis) asymptotically (漸近地). (d) The flatness (平坦程度) of the curve is determined by and the position (位置) the curve is determined by . (e) The total area under the normal curve is 1. (f) The area under the curve from x = a to x = b represents the proportion (probability) of data (數據) for x takes up values from a to b, or in symbolic notation P( a x b ) (g) The probability that x takes up a single value (某一數值) is zero (零). When x is continues. F7-MS-Ch16 - 1 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications (h) F7-MS-Ch16 - 2 The probability that x takes up a given standard deviation within the mean is a constant (在平均值內某指定標準差一常數). Suppose x ~ N( , 2 ), then (i) P( x ) =0.6826 (ii) P( 2 x 2 ) =0.9544 (iii) P( 3 x 3 ) =0.9974 § 16.3 Application of normal distribution 正態分佈的應用 Common application of normal distribution are (a) height (b) mass (c) examination results (d) I.Q. score F7-MS-Ch16 - 2 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications F7-MS-Ch16 - 3 § 16.3.1 The standard normal distribution 標準正態分佈 If x ~ N (0,1), then x is a standard normal variable (標準正態變量) or x follows (依循) a standard normal distribution. The p.d.f. f(x) of standard normal distribution is f(x)= § 16.3.2 Standard normal table and standardization (a) 1 2 e x2 2 標準正態分佈表與標準化 Standard normal table 標準正態分佈表 If z ~ N (0, 1), then the probability that z lies within the range from 0 to k, is given by P( 0 z k ) = area under the curve from x = 0 to x = k, and the value may be easily found from the standard normal table. (b) Standardization 標準化 If x ~ N( , 2 ), then we may standardize x to a standard normal variable z by letting z then z ~ N(0,1). Example 1 If z is a standard normal variable (標準正態變量),evaluate the following (計算下列): (a) P( 0 z 1.2 ) = 0.3849 (to 4 d.p) F7-MS-Ch16 - 3 x , F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications (b) P( 0 z 2.54) = 0.4945 (to 4 d.p) (c) P( 0.5 z 0 ) = 0.1915 (to 4 d.p) (d) P( 1.39 z 0 ) = 0.4177 (to 4 d.p) (e) P( 1.1 z 1.78 ) = 0.4625 – 0.3643 = 0.0982 F7-MS-Ch16 - 4 F7-MS-Ch16 - 4 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications (f) P( 056 z 0.47 ) = 0.2123 + 0.1808 = 0.3931 (to 4 d.p) (g) P( 1.86 z 0.11 ) = 0.4686 – 0.0438 =0.4248 (h) P( 0.32 z ) = 0.1255 +0.5 =0.6255 (to 4 d.p) F7-MS-Ch16 - 5 F7-MS-Ch16 - 5 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications (i) P( z 0.69 ) = 0.5 – P( 0.69 z 0) = 0.5 – 0.2549 = 0.2451 (to 4 d.p) (j) P( z 0.61) = 0.5 – P( 0 z 0.61) = 0.5 – 0.2291 = 0.2709 (k) P( z 1.66 ) = 0.5 + 0.4515 = 0.9515 F7-MS-Ch16 - 6 F7-MS-Ch16 - 6 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications F7-MS-Ch16 - 7 Example 2 If z ~ N(0,1), which of the following values P1,P2,P3,P4 is the greatest (最大)? (a) P1 = P( 0 z 1) = 0.3413 (b) P2 = P( 1 z 2 ) = 0.4772 – 0.3413 = 0.1359 (c) P3 = P( 0.5 z 0.5 ) = (0.1915)2 = 0.383 (d) P4 = P( 0.4 z 0.6 ) = 0.1554 + 0.2257 = 0.3811 F7-MS-Ch16 - 7 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications Example 3 If z ~N(0,1),find the value of k for each of the following : (a) P( 0 z k ) = 0.3289 k 0 0.95 1 k =0.95 (b) P( 0 z k ) = 0.4920 k = 2.41 (c) P( k z 0 ) = 0.3531 –k = 1.05 k = –1.05 (d) P( k z 0 ) = 0.1064 k = –0.27 (e) P( k z 0.2 ) = 0.2347 P(k<z<0) = 0.2347 – 0.0793 –k = 0.4 k = –0.4 (f) P( k z 1.23 ) = 0.0912 0.3907 – P( 0 z k ) = 0.0912 P( 0 z k ) = 0.2995 k = 0.84 (g) P( 1 z k ) = 0.5536 P( 0 z k ) = 0.5536 – 0.3413 P( 0 z k ) = 0.2123 k = 0.556 F7-MS-Ch16 - 8 F7-MS-Ch16 - 8 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications (h) P(-1.4<z<k) = 0.2207 P(k<z<0) = 0.4192 – 0.2207 P(k<z<0) = 0.1985 k = – 0.52 (i) P(z>k) = 0.0668 P( 0 z k ) = 0.5 – 0.0668 k = 1.50 (j) P(z<k) = 0.2514 P(k<z<0) = 0.5 – 0.2514 P(k<z<0) = 0.2486 k = – 0.67 (k) P( z k ) = 0.9292 P(–k<z<0) = 0.9292 – 0.5 P(–k<z<0) = 0.4292 k = – 1.47 (l) P(z<k) = 0.6217 P( 0 z k ) = 0.1217 k = 0.31 Example 4 If x ~ N(24, 9) evaluate the following : (a) P( 23.1 x 26.7 ) = P( 23.1 24 26.7 24 z ) 3 3 = P( 0.3 z 0.9 ) = 0.1179 + 0.3159 = 0.4338 (b) P(22.5<x<25.8) =P( 22.5 24 25.8 24 z ) 3 3 F7-MS-Ch16 - 9 F7-MS-Ch16 - 9 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications =P( 0.5 z 0.6 ) = 0.1915 + 0.2257 = 0.4172 (c) P( x 28.2 ) = P( z 28.2 24 ) 3 = P( z 1.4 ) = 0.5–0.4192 = 0.0808 (d) P( x 27.3 ) = P( z 27.3 24 ) 3 = P( z 1.1 ) = 0.5 + 0.3643 = 0.8643 Example 5 If x ~ N(34, 25), evaluate (a) P( 28 x 32 ) = P( 28 34 32 34 z ) 5 5 = P( 1.2 z 0.4 ) = 0.3849 – 0.1554 = 0.2295 (b) P( x 37.1) P( z 37.1 34 ) 5 = P( z 0.62 ) = 0.5 – 0.2324 = 0.2676 (c) P( x 26.8 ) P( z 26.8 34 ) 5 = P( z 1.44 ) = 0.5 – 0.4251 = 0.0749 F7-MS-Ch16 - 10 F7-MS-Ch16 - 10 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications (d) P( x 30 | x 26.8) 30 34 26.8 34 |z ) 5 5 = P( z 0.8 | z 1.44 ) = P( z =P = 0.4251 0.2881 0.5 0.4251 0.137 0.9251 = 0.1481 (to 4 d.p) Example 6 If x ~ N(50, 100), find the values of k if (a) P( 46 x k ) = 0.6530 P( 46 50 k 50 z ) = 0.6530 10 10 P( 0.4 z k 50 ) = 0.6530 10 k 50 = 2.82 10 k = 78.2 (b) P( k x 52 ) = 0.2881 P( k 50 52 50 z ) = 0.2881 10 10 P( k 50 z 0.2 ) = 0.2881 10 k 50 = –0.55 10 k = 44.5 P( k 50 52 50 z ) = 0.2881 10 10 P( k 50 z 0 ) = 0.2088 10 k 50 = –0.55 10 k = 44.5 F7-MS-Ch16 - 11 F7-MS-Ch16 - 11 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications Example 7 If x ~ N(12, 4), find the value of k if P( 7 x k ) = 0.8176 P( 7 x k ) = 0.8176 P( 7 12 k 12 z ) = 0.8176 2 2 P( 2.5 z k 12 ) = 0.8176 2 k 12 = 0.93 2 k = 13.86 Example 8 If z ~ N(0, 1). solve k for the following to 4 decimal places: (a) P( 0 z k ) = 0.4 k = 1.28 + 0.01( 0.4000 0.3997 ) 0.4015 0.3997 k = 1.28 + 0.01( 3 ) 18 = 1.2817 (to 4 d.p) (b) P( 0 z k ) = 0.25 P( 0 z k ) = 0.67 + 0.01( 14 ) 31 = 0.6745 (to 4 d.p) (c) P( k z 0 ) = 0.365 k = -1.10 + ( = -1.1032 7 )0.01 22 (to 4 d.p) (d) P( k z 0 ) = 0.4324 k = -1.49 + 0.01( 5 ) 13 = -1.4938 (to 4 d.p) (e) P( 0 z k ) = 0.1057 k = 0.26 + 0.01( 31 ) 38 = 0.2682 (to 4 d.p) F7-MS-Ch16 - 12 F7-MS-Ch16 - 12 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications § 16.4 Problems involving normal distribution F7-MS-Ch16 - 13 正態分佈應用題 Example 9 The lengths of sardine(沙甸魚) follow (依循) a normal distribution of mean 10cm and standard deviation 1.2 cm. If a sardine can (罐) contains sardine with lenghts range (範圍) from 9.4cm to 10.9cm, find the probability that a randomly selected sardine are unsuitable (不適合) for can food. Let x be the length of sardine x ~ N(10, 1.22) P(suitable) = P( 9.4 x 10.9 ) = P( 9.4 10 10.9 10 z ) 1.2 1.2 = P( 0.5 z 0.75 ) = 0.1915 + 0.2734 = 0.4649 P(unsuitale) = 1 – 0.4649 = 0.5351 Example 10 The amount of soft drink (飲料) a machine discharges (排出) follow a normal distribution with mean 8.oz. and standard deviation 0.2 oz. If the capacity (容量) of the cup is 8.28 oz., (a) find the probability that a particular cup of soft drink overflows (滿溢). Let X of be amout of soft drink discharges per cup x ~ N(8, 0.22) P( x 8.28 ) =P(z 8.28 8 ) 0.2 = P( z 1.4 ) = 0.5 – 0.4192 = 0.0808 (b) find the probability that 2 out of the 5 cups of soft drink discharged by the machine overflows. Reqired probability = C52(0.0808)2 (0.9192)3 = 0.0507 (to 4d.p) (c) find the probability that the 2nd overflow found on the 5th discharge. Required probability = (0.0808)[C4(0.0808)(0.9192)3] = 0.0203 (to 4 d.p) (d) find the adjusted (調節) capacity of the cup so that only 1% of the cups overflow. k = adjusted capacity of the cup P(x>k) = 0.01 F7-MS-Ch16 - 13 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications P( z F7-MS-Ch16 - 14 k 8 ) =0.49 0.2 k 8 2.3267 0.2 k = 8.4653 (to 4d.p) Example 11 The amount of chemiacl (化學用品) used per week in a certain factory is a normal variable (正 態變量) with mean 1800 kg and standard deviation (標準變量) with mean 1800 kg and standard deviation (標準差) 200kg. If the chemical used in the factory orders once a week (一週落一次 單), find the minium stock (存貨) the company should order at the start of a week to ensure that there is at most a probability of 0.0017 of running out of stock (不夠存貨). Let x be the amount of chemical used per week x ~ N (1800, 2002) k = order per week P( x k ) 0.0017 P( 0 z k 1800 ) 0.4983 200 k 1800 2.93 200 k 2386 The minium stock order is 2386 kg Example 12 The liferime of a particular type of batteries (電池) follows a normal distribution with mean 1500 hours and a standard deviation of 150 hours. (a) Find the interquartile range (四分位數間距) of the lifetime of the batteries. (b) Find the proportion (比例) of batteries with lifetime at least (不少於) 1380 hours. F7-MS-Ch16 - 14 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications F7-MS-Ch16 - 15 (c) Suppose a machine is powered (發動) by 3 of such batteries, and the machine works (工作) if at least 2 of the three batteries with lifetimes at least 1380 hours. Find the probability that the machine works. (d) Given that the machine works, find the probability that all the three batteries with lifetimes at least 1380 hours. F7-MS-Ch16 - 15 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications F7-MS-Ch16 - 16 Example 13 The number of customers served (服務) in a particular restaurant daily is a normal varibale (正態 變量) with mean 156 and standard deviation (標準差) of 18. Staffs receive a special payment (花 紅) if the number of customers in that day exceeds (超越) 185. Find the probabilites that (a) staffs will receive a special payment in a particular day. x: no. of customers served per day x ~ N(156,182) P( x 185.5 ) 185.5 156 ) 18 = 0.5 P(0 z 1.64) = 0.5 P(0 z = 0.5 – 0.4495 = 0.0505 (b) the number of customers served in a particular day lies between 149.5 and 164.5 inclusively (包括在內). P (149.5 x 164.5) 149.5 156 164.5 156 z ) = P( 18 18 = P(0.36 z 0.47) = 0.1406 + 0.1808 = 0.3214 (c) the fourth day of a particular week is the first day that staffs will receive a special payment. Required probability = (0.9495)3 (0.0505) = 0.0432 (to 4 d.p) F7-MS-Ch16 - 16 F7 Mathematics and Statistics Chapter 16 The Normal Distribution and Its Applications F7-MS-Ch16 - 17 Example 14 The number of words per page in a particular book is a normal variable with mean 700 and standard deviation of 10 words. The book contains 250 pages. (a) Find the probability that the number of words in a particular page lies between 695 and 710 inclusively. (b) Find the probability that the fifth page(第五頁) is the second page with number of words between 675 and 710 inclusively. (c) Find the expected number of pages with number of words between 695 and 710 inclusively. (d) Find the probability that exactly 2 of the first 5 pages with number of words between 695 and 710 inclusively. F7-MS-Ch16 - 17