STAT 400 Discussion 09 Solutions Spring 2018 - GitHub Pages

1 STAT 400 Discussion 09 Solutions Spring 2018 1. Consider a 6-pack of soda. Suppose that the amount of soda in each can follows a normal distribution with mean oz and standard deviation oz. Assume that all cans are filled independently of each other. Find the probability of the following: = , s = a) a can is underfilled, there is less than 12 oz of soda in a can; P(X < 12) = P = P(Z < ) = F ( ) = b) all 6 cans are underfilled; ( ) 6 = OR 6 C 6 ( ) 6 ( ) 0 = c) at least one of the 6 cans is underfilled; 1 P(none) = 1 ( ) 6 = d) exactly 2 of the 6 cans are underfilled; 6 C 2 ( ) 2 ( ) 4 = e) the average amount of soda in these 6 cans is less than 12 oz. Need P( < 12) = ? n = 6 Normal distribution. P( < 12) = P = P(Z < ) = F ( ) = -< -< 2. The weights of the eggs at a particular farm are normally distributed with the mean weight of oz and standard deviation oz.

2 A) Find the probability that a randomly selected egg weighs over oz. = , s = P ( X > ) = P = P ( Z > - ) = b) If three dozen eggs (36 eggs) are randomly and independently selected, what is the probability that exactly 30 of them weigh over oz? Let W = the number of egg (out of 36) that weigh over oz. Eggs are selected independently Binomial distribution, n = 36, p = ( see part (a) ). Need P(W = 30) = ? P(W = 30) = = = c) What is the probability that the total weight of (randomly and independently selected) three dozen eggs (36 eggs) is over 60 oz? n = 36. Need P( Total > 60 ) = ? We sample from a normally distributed population. The total weight of (randomly and independently selected) three dozen eggs is normally distributed with mean 36 1 egg = 36 = 63 oz and standard deviation = oz. z = = P( Total > 60 ) = P( Z > ) = -> )1( )P(630)15870()84130( 630)15870()84130(792,947, egg 1 = Let X equal the weight in grams of a miniature candy bar.

3 Assume that = E ( X ) = and s 2 = Var ( X ) = Let be the sample mean of a random sample of n = 30 candy bars. Find: (a) E ( ). (b) Var ( ). (c) P ( < < ), approximately. 4. Approximate P ( < < ), where is the mean of a random sample of size n = 32 from a distribution with mean = 40 and variance s 2 = 8. XXXXXX5. The weight of an almond varies with mean ounce and standard deviation ounce. a) What is the probability (approximately) that the total weight (of a random sample) of 64 almonds is greater than 3 ounces? E( Total ) = 64 = Var( Total ) = 64 2 = SD( Total ) = n = 64 large. Total is approximately normally distributed. P ( Total > 3 ) P = P ( Z > ) = b) Determine the sample size (the number of almonds) needed to have the probability of at least that the total weight is greater than 16 ounces. P( Z > ) = Z.

4 N 16 0. = or 2 n = at least 343. ( round up ) Indeed, n = 342, P ( Total > 3 ) P P ( Z > ) = n = 343, P ( Total > 3 ) P P ( Z > ) = -> Total -nnn n + -> -> 6. An instructor gives a test to a class containing several hundred students. It is known that the standard deviation of the scores is 14 points. A random sample of 49 scores is obtained. = ? , s = 14, n = 49. a) What is the probability that the average score of the students in the sample will differ from the overall average by more than 2 points? Need 1 P( 2 < < + 2 ) = ? n = 49 large . Central Limit Theorem: . 1 P( 2 < < + 2 ) = 1 P( < Z < ) = b) What is the probability that the average score of the students in the sample will be within 3 points of the overall average? Need P( 3 < < + 3 ) = ?

5 P( 3 < < + 3 ) = P( < Z < ) = XZX =-n X()() -+<<---49142Z49142P1 XX()() -+<<--49143Z49143P 7. Let X 1 and X 2 be independent with normal distributions N ( 6, 1 ) and N ( 7, 1 ), respectively. Find P ( X 1 > X 2 ). Hint: Write P ( X 1 > X 2 ) = P ( X 1 X 2 > 0 ) and determine the distribution of X 1 X 2 . E ( X 1 X 2 ) = E ( X 1 ) E ( X 2 ) = 1. Var ( X 1 X 2 ) = Var ( X 1 ) + Var ( X 2 ) = 2. X 1 X 2 has Normal distribution. P ( X 1 > X 2 ) = P ( X 1 X 2 > 0 ) = = P ( Z > ) 8. Compute P ( X 1 + 2 X 2 2 X 3 > 7 ), if X 1 , X 2 , X 3 are with common distribution N ( 1, 4 ). E ( X 1 + 2 X 2 2 X 3 ) = E ( X 1 ) + 2 E ( X 2 ) 2 E ( X 3 ) = 1. Var ( X 1 + 2 X 2 2 X 3 ) = Var ( X 1 ) + 4 Var ( X 2 ) + 4 Var ( X 3 ) = 36. SD ( X 1 + 2 X 2 2 X 3 ) = 6. X 1 + 2 X 2 2 X 3 has Normal distribution. P ( X 1 + 2 X 2 2 X 3 > 7 ) = = P ( Z > ) = () --> 210 ZP -> 617 ZP 9.

6 Let X 1 , X 2 , .. , X 70 be a random sample of size n = 70 from a distribution with f ( x ) = ( 6 x ) 2, 0 < x < 6, zero elsewhere. Find P ( < ) approximately. E ( X ) = = E ( X 2 ) = = Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 2 = n = 70 large. Z. P ( < ) = P ( Z < ) = 721X()() - =-602 6721 dxxxdxxfx ()() - =-60222 6721 dxxxdxxfx Xn-X -< 10. A machine operation produces widgets whose diameters are normally distributed, with a mean of inches and a standard deviation of inches. Suppose that specifications specifications require that the widget diameter be inches plus or minus inches ( that is, between and inches ). a) What proportion of the production will be unacceptable? P ( Acceptable ) = P ( < X < ) = = P ( < Z < ) = = P ( Unacceptable ) = 1 P ( Acceptable ) = 1 = b) Suppose 25 widgets are independently and randomly selected from the production process.

7 What is the probability that exactly 2 of the 25 will be unacceptable? 25 C 2 2 23 = ( Binomial Distribution ) c) A quality control inspector selects 25 widgets from the production independently and at random. If the average diameter of the selected widgets is within inches of inches ( that is, between and inches ), the production process is allowed to continue. However, if the average diameter of the selected widgets is not within inches of inches, the production process is stopped and the machine is checked. What is the probability that the production process will be stopped after examining a random sample of 25 widgets? n = 25. Need 1 P( < < ) = ? We sample from a Normal distribution. P ( < < ) = = P ( < Z < ) = = 1 = -<< XXZ-= -<< 11. 10. (continued) After a disgruntled employee kicked the machine, the mean shifted to inches, and the standard deviation shifted to inches, while the distribution of the diameters remained normal.

8 A) What proportion of the production will be unacceptable after the machine is kicked? P ( Acceptable ) = P ( < X < ) = = P ( < Z < ) = = P ( Unacceptable ) = 1 P ( Acceptable ) = 1 = b) After the machine was kicked, 25 widgets are independently and randomly selected from the production process. What is the probability that at least 7 of the 25 will be unacceptable? Using cumulative Binomial probabilities table or a computer: P ( X 7 ) = 1 P ( X 6 ) = 1 CDF @ 6 = 1 = c) After the machine was kicked, what is the probability that the production process will be stopped after examining a random sample of 25 widgets? n = 25. Need 1 P( < < ) = ? We sample from a Normal distribution. P ( < < ) = = P ( < Z < ) = = 1 = -<< XXZ-= -<< 12. Let X 1 , X 2 , .. , X n be a random sample from: N ( , s 2 ) unknown, s known.

9 Show that = is the MLE for .. ln = . = = 0. = n = 0. = = . X()() = --=niinxxxx122 221 21 21,..,,; exp L() L ()()() =----niixnn122 21 2 2lnln() L ddln() =-niix12 1() 1 niix=- 1 niix= =niin1 X 1X13. Let X 1 , X 2 , .. , X n be a random sample from: N ( , s 2 ) known, s unknown. Show that is the MLE for s 2.. ln = . = = 0.. OR = . q = s 2 ln = . = = 0.. () =-=niin122 X 1 ()() = --=niinxxxx122 2212 21 21,..,,; exp L() L ()()() =----niixnn122 21 2 2lnln() L ddln() =-+-niixn123 1() =-=niin122 X 1 ()() = --=niinxxxx122 2212 21 21,..,,; exp L() -- = 21 21 12 exp niinx() L ()()() =----niixnn12 21 22 2lnln() L ddln() =-+-niixn122 212() =-==niin122 X 1 14.

10 Let q > 0 and let X 1 , X 2 , .. , X n be a random sample from the distribution with the probability density function , x > 0. a) Find the method of moments estimator of q. E ( X ) = y = E ( X ) = Integration by parts: Choice of u: L ogarithmic A lgebraic T rigonometric E xponential u = y 2, dv = , du = 2 y dy, v = . E ( X ) = = u = y, dv = , du = dy, v = . E ( X ) = = ()() XX 2 ;xexxfxf-== ~() - =0 0 2 dxxxdxxfxxe x 2xdxdy= - 0 2 dyyye -=babaduvabvudvu dyye - ye -- --- +0 2 20 dyyyyeey - 0 2dyyyedyye -ye 1- - - -- +0 10 1 2dyyyeey - 0 2dyye = = . OR E ( X ) = y = E ( X ) = = E ( Y 2 ), where Y has Exponential distribution with mean . E ( X ) = E ( Y 2 ) = Var ( Y ) + [ E ( Y ) ] 2 = + =.