M4 Problem Set

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Jan 9, 2024

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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‘
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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
(Remember: You may find the standard normal table by clicking Help Files.) Be sure to number each of your answers. 1. Find P(Z <1.93) 2. Find P(Z <« —.05) Your Answer: 1. Find P(Z = 1.93). From the table, we get P(Z < 1.93) = .97320 2. Find P(Z < - .05). From the table, we get P(Z < - .05) = .48006 1. Find P(Z = 1.93). From the table, we get P(Z< 1.93) = .97320. 2. Find P(Z = - .05). From the table, we get P(Z<-.05) = .48006. Question 5 0/0 pts (Remember: You may find the standard normal table by clicking Help Files)
Be sure to number each of your answers. 1. Find P(Z = 2.08) 2.Find P(Z = —.78) 3.Find F(—.220 < 7 £ — G3) Your Answer: 1. P(Z < 2.006) is given in the standard normal table .98030 So, P(Z 22.06) = 1 - P (Z < 2.06)=1 -.98030 =.0197 2. P(Z< -.78). is given in the standard normal table .21770. So,P(Z=-.78)=1-P(Z<-.78)=1-.21770 = .7823. 3. We have P(-2.20 < Z < -.63) = P(Z < -.63) - P(Z < -2.20). The values in the table: P(-2.20 < Z <-.63) =.26435 - .01390 = .25045
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1. Find P(Z 2 2.06). P(Z < 2.06). is given directly in the standard normal table and is found to be .98030. So, P(Z22.06)=1-P(Z< 2.06) =1 -.98030 =.0197 2. Find P(Z 2 -.78). P(Z=< -.78). is given directly in the standard normal table and is found to be .21770. So, P(Z2-.78)=1-P(Z<-.78)=1-.21770 = .7823. 3. Find P(-2.20 = Z< -.63). Again, we have: P(-2.20= Z £-.63)=P(Z =< -.63) - P(Z =< -2.20). Reading the values in the table: P(-2.20 = Z < -.63) =.26435 - .01390 = .25045 Question 6 0/0 pts A farmer harvests several heads of iceberg lettuce. The weight of the heads of lettuce are normally distributed with a mean weight of 2.2 pounds and a standard deviation of .55 pounds. If you choose a head of lettuce at random, what is the probability that the head you chose will weigh: a) Less than 2 pounds? b) Greater than 2.5 pounds? c) Between 1.8 pounds and 2.7 pounds? Your Answer:
a) z score for x=2: 2—-2.2 From the table P(Z < -.36) = .35942 b) z score for x=2.5: 25-2.2 55 59 P(Z2.55)=1.0-P(Z<.55)=1-.70884=.29116 c) z score for x=2.7: 2.7-2.2 5 — 91 z score for 1.8: 1.8-2.2 _ o (9 P(-173<Z<.91)=P(Z<.91)— P(Z< —.73) P(—.73< Z <.91) = .81859 — .23270 = .58589
We have y=2.2 and 0=.55. a) We must find the z-score for x=2: x—u 2-22 a .55 = —-.36. - - So, we want P(Z < -.36) From the table, we find. P(Z <-.36) = 35942, b) We must find the z-score for x=2.5: _x—p 25-22 _ =% T~ 5 55 So, we want P(Z = -.55). Since this is greater than, we must use: P(Z =2.55)=10—-P(Z =<.55)=1-.70884 = .29116. c) We must find the z-score for x=2.7: _x—p 27-22 o .55 o Z and the z-score for x=1.8: So, we want P(-73 < Z < 91) P(-73 £Z < 91)=P(Z< 91)-P(Z < -73). P(-73 < Z = .91) =.81859 —.23270 = .58589. Question 7 0/0pts The time that people stand in the waiting line at a particular fast food restaurant is normally distributed with a mean time of 130 seconds and a standard deviation of 25 seconds. If you go to that fast food restaurant, what is the probability that you will stand in the waiting line for:
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a) More than 150 seconds? b) Less than 120 seconds? c) Between 110 seconds and 150 seconds? Your Answer: a) z score for x=150 150-130 __ = =.8 P(Z> .80)=1.0— P(Z < .80) = 1—.78814 = .21186 b) z score for x=120 120-130 _ w— = —4 P(Z < .4) = .34458 c) z score for x=150 150—130 = — -8 z score for x=110 110-130 _ —— =8 P(-8<Z<.8)=P(Z<.8)—P(Z<-2.8) P(—.8< Z < .8) =.78814 — .21186 = .57628
We have p=130 and 0=25. a) We must find the z-score for x=150: x—p 150-130 ” i - So, we want P(Z =2 .80) Since this is greater than, we must use: P(Z =2.80)=10-P(Z <.80)=1-.78814 =.21186. b) We must find the z-score for x=120: x—p 120-130 o 25 N - - So, we want P(Z < -.4). From the table, we find. P(Z<-.4) = .34458. c) We must find the z-score for x=150: __x—u _150-130 _ . . g 25 = and the z-score for x=110: _x—p 110-130 _ =g e g & b4 So, we want P(-8<Z < .8) P(-8<Z < 8)=P(Z< 8)—-P(Z < -8). P(-8 < Z < .8)=.78814 —.21186 = .57628. Question 8 5/5 pts As a reminder, the questions in this review quiz are a requirement of the course and the best way to prepare for the module exam. Did you complete all questions in their entirety and show your work? Your Answer:
yes Quiz Score: 5 out of 5
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