M4 Problem Set

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Statistics

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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 5 out of 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 2 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) = a? = (= — p)’ £ (a)) (—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) = ¢* = Z(x — pu)*f(x) = = (0—2.46)*(.15) + (1—2.46)%(. 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 = V2.208% = 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 )|p (1 _p)(n—w) 13! 4 (13—4) Ty 2114(1 — .211) _ ,168‘ Solution. Here we have n=13, x=4, and p=.211. flx) = 13! 17 ¥(1 —p)in—= = .211%(1 —.211)(33-4) = 168, x'(n 41(13 - 4)!" 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: For two defective, we have n=6, x=2, p=.17 2 (6-2) ‘1,(6 A7 (11— 17) ) = 2057‘
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For one defective, we have n=6, x=1, p=.17 6! 1 (6—-1) a1 anC = .40178‘ For zero defective, we have n=6, x=0, and p=.17 6! 0 (6—0) ‘0!(6_0)!.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. .173(1 — .17)(6-2) = 2057. 6! FlE) = 21(6 — 2)] ! (nn-.- x)! p*(1 —p)n=¥ = For one defective, we have n=6, x=1, and p=.17. (] — p)n=) = - 174(1 - 17)(6Y = 40178, ! 11(6-1) n! I = xl(n - x)!p For zero defective, we have n=6, x=0, and p=.17. flx) = ¥(1—p)n—%) = .17%(1 — .17)¢% = 3269. n! ! Xn=-x" 01(6 — 0)! 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 = 2.086) 2.Find P(Z = —.78) 3.Find F(—.220 < 7 £ — .63) 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: 2.5-2.2 55— 99 P(22.55)=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 - P(-73<Z<.91)=P(Z<.91)—P(Z<—.73) P(-.73< Z <.91) = .81859 — .23270 = .58589
We have p=2.2 and 0=.55. a) We must find the z-score for x=2: Xx—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 = =5 55 > A— - So, we want P(Z 2 -.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 e 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 0/0 pts 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 _ 25 = .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 _ — % S z score for x=110 110-130 _ = —8 P(-8<Z<.8)=P(Z<.8)—P(Z<-.8) P(—8<7Z<.8)=.78814 — .21186 = .57628
We have y=130 and 0=25. a) We must find the z-score for x=150: x—pu 150-130 5 i So, we want P(Z = .80) Since this is greater than, we must use: P(Z >.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 2 T e 25 = and the z-score for x=110: x—p _110-130 _ g 25 > - — - — 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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