M4 Problem Set

M4: Problem Set Due No due date Points 5 Questions 8 Time Limit None Attempt History Attempt Time Score LATEST Attempt 1 229 minutes 5outof 5 Score for this quiz: 5 out of 5 Submitted Aug 10 at 1:24pm This attempt took 229 minutes. Question 1 0/0 pts A barber expects to get between zero and six customers per hour in his barber shop. The probability of these is given as follows: Customers Probability, f(x) 0 A5 1 .07 2 29 3 .26 4 A3 5 .09 6 01 Find the number of expected customers that the barber will get per hour. Also, find the variance and standard deviation of this data. Your Answer: Number of expected customers per hour = 2.46 E(z) =p=Xzf(z) (.15) +1(.07) +2(.29) +3(.26) +4(.13) +5(.09) + 6 (.01) =

Variance = 2.2084 Var (2) = o® = 5(z — )’ f ()| (—2.46)* (.15) + (1 — 2.46)* (.07) + (2 — 2.46)* (.29) + (3 — 2.4 1(.09) + (6 — 2.46)” (.01) = 2.2084| Standard Deviation = 1.486 o=V 1/2.:2084 = 1.486) Solution. The expected value is given by E(x)=u=Zxf(x) = =0(.15) + 1(.07) + 2(.29) + 3(.26) + 4(.13) + 5(.09) + 6(.01) = 2.46. So, the barber can expect 2.46 customers per hour. The variance is given by Var(x) = 6* = L(x — p)*f(x) = = (0—2.46)*(.15) + (1—-2.46)3(.07) + (2—2.46)%(. 29) + (3—2.46)*(. 26) + (4—2.46)*(. 13) + (5-2.46)*(.09) + (6—2.46)*(.01) = 2.2084. The standard deviation is given by: o = o2 =+/2.2084 = 1.486. \/_ Question 2 0/0 pts

A baseball player has a batting average of .211 (in other words, he gets a hit 21.1 % of the time that he goes up to bat). If he goes up to bat 13 times, what is the probability that he will get exactly 4 hits? Your Answer: Here we have n=13, x=4, p=.211 ‘f(w) = P ( -p)"® 13! 4 (13—4) _ 2114 (1 - 211) Y = .168‘ Solution. Here we have n=13, x=4, and p=.211. 13 x — (n—-x) - X - _— . (13-4) — : P A -p) o A2 168 flx) = x'(n- Question 3 0/0 pts A large shipment of light bulbs has just arrived at a store. It has been revealed that 17 % of the light bulbs are defective (the other light bulbs are good). Suppose that you choose 6 light bulbs at random. What is the probability that 2 or less of the bulbs are defective. Your Answer: ‘f (2) = sgp" (L= )" For two defective, we have n=6, x=2, p=.17 6! 2 (6-2) ‘1!(6_2)!.17 (1—.17)62) — .2057‘

For one defective, we have n=6, x=1, p=.17 6! 1 (6—-1) ‘1!(6_1)!.17 (1—.17)6D = .40178‘ For zero defective, we have n=6, x=0, and p=.17 6! 0 (6—0) ‘O!(G_O)!.17 (1—.17)69 — 3269 Solution. For two or less, we must calculate the probability of getting two defective, one defective, and zero defective, then add the probabilities. For two defective, we have n=6, x=2, and p=.17. 6! . *(1—p)>) = 173(1 — 17)3) = 2057. n HE) =4 ~21(6 = 2)] 1 (n-—x)!p For one defective, we have n=6, x=1, and p=.17. n! 6! T = Jc)‘p“’(l —p)n=¥) = 1731 - A7)V = 40178. T3> 116 - 1! For zero defective, we have n=6, x=0, and p=.17. n! 6! x — (n-x) = L 0 -_— 7 (6—-0) =3 A n_x)!p (1-p) 0T(6 — 0)] 17°(1 - .17) 3269 f(x)=x!( So, the probability of two, one, or zero defective: .2057+.40178+.3269=.93438. Question 4 0/0 pts