M4 Exam

M4: Exam - Requires Respondus LockDown Browser + Webcam Due No due date Points 50 Questions 5 Time Limit 90 Minutes Requires Respondus LockDown Browser Instructions You may only have the following items when taking an exam: computer, 1-2 pieces of blank scratch paper, a pen/pencil, and a calculator. You may ONLY use the equation sheets that are provided WITHIN the exam. The use of printed versions will be considered a violation of the Academic Integrity Policy. Attempt History Attempt Time Score LATEST Attempt 1 90 minutes 43 out of 50 Score for this quiz: 43 out of 50 Submitted Aug 14 at 4:30pm This attempt took 90 minutes. Question 1 717 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet =

A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency? Your Answer: For six; n=8, x=6, p=.08 ‘f (2) = stp” (1 =0) """ = ey - 08°(1 - .08)°™" =6. For six; n=8, x=7, p=.08 @) = g (L -p) ") = gy - 087(1 - 08)° 7 =1 For six; n=8, x=8, p=.08 n! T n—x ! 8—8 f(@) = 525 (1 -p)"" = gy - -08°(1-.08) = 1. The probability of six or more safety systems failing: 6.2 x 107° +1.54 x 107" +1.68 x 107? = 6.355 x 107° « G b For six, we have n=8, x=6, and p=.08. n! f&x) =7 = p*(1—p)n=x) = 085(1 —.08)(®-¢) = 6.2 x 1075, 8! 6! (8 —6)!" For six, we have n=8, x=7, and p=.08. n! 8! o x(1 — ) (n=x) _ : 701 — (8-7) — -7 f(x) T Yy p*(1—p) @ =7 087(1—.08) 1.54x 107", For six, we have n=8, x=8, and p=.08. n! s 8 B S ; f(x) = pry e (1—p)n—=) = T 8)!.083(1 .08)®-%) = 168 x 107", So, the probability of six, seven, or eight failing: 6.2x107%+ 1.54x1077 +1.68x107° = 6.355x 107°. Question 2 12 /12 pts

d. P(Z < -1.21)= .11314. b. P(Z 2 -1.76)=1- .03920= .9608. C. P(1.63 < Z < 2.36)= .99086- .94845= .04241. Question 3 8 /12 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & Standard Normal Table & (https://iprevious.nursingabc.com/upload/images/Help_file picture/standardn A company manufactures a large number of rods. The lengths of the rods are normally distributed with a mean length of 5.1 inches and a standard deviation of .8 inches. If you choose a rod at random, what is the probability that the rod you chose will be: a) Less than 4.5 inches? b) Greater than 5.0 inches? c) Between 4 inches and 5.3 inches? - 4 Your Answer: a.

We have and Must find Z-score for x=4.5 T—H 4.5—5.1 - = —.75‘ ‘z: — p— We want‘P (Z < —.75)‘ From table we find ‘P (Z < —.75)‘=.22965 b. Must find Z-scoe for x=5 ‘Z: m;u _ 4.0;5.1 _ _1.375‘ We want \P(Z > —1.375) = 1.0 — P(Z < —1.375) = 1 — .21186 = .78814 C. Z-score for 4 z= T = 25— 375 4 ) >

Question 4 10/ 12 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet 5 (https://previous.nursingabc.com/upload/images/Help_file_picture/Statistics Standard Normal Table & (https://previous.nursingabc.com/upload/images/Help_file_picture/standardn A life insurance salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as follows: Policies Sold Probability, f(x) Per Day 0 .08 1 14 2 16 3 25 4 24 5 13 Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data. - J 4 Your Answer: B(2) = = 5af () (.08)+1(.14)+2(.16)+3(.25)+4(.24)+5(.13)=.08+.14+.32+.75+.96+.65=3.62 Can sell 3.62 per day Variance Var (z) = o® = S(z — p)’ f (x)]

(0 — 3.62)" (.08) + (1 — 3.62)* (.14) + (2 — 3.62)” (.16) + (3 — 3 Standard deviation o = Vo? = 1/3.62 = 1.903| The expected value is given by E(x)=p=Zxf(x)= =0(.08)+1(.14)+2(.16)+3(.25)+4(.24)+5(.13)=2.82. So, the insurance salesperson can expect to sell 2.82 policies per day. The variance iIs given by Var(x) = ¢* = Z(x — u)*f(x) = = (0 — 2.82)%(.08) + (1 — 2.82)%(. 14) + (2 — 2.82)%(. 16) + (3 — 2.82)*(.25) + (4 — 2.82)%(. 24) + (5 — 2.82)*(.13) = 2.1676. The standard deviation is given by: g =2 =v2.1676 = 1.4722, Question 5 6/7 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & (https:/Iprevious.nursingabc.com/upload/images/Help_file picture/Statistics Standard Normal Table = (https://previous.nursingabc.com/upload/images/Help_file picture/standardn An archer is shooting arrows at a target. She hits the target 77% of the time. If she takes 20 shots at the target, what is the probability that she