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This attempt took 84 minutes.
Question 1
7 / 7 pts
You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) In a large shipment of clocks, it has been discovered that 15 % of the clocks are defective. Suppose that you choose 9 clocks at random. What is the probability exactly 2 of the clocks are defective. Your Answer: n=9 x=2 p=. f(x)= n! x! ( n − x )! px(1-p)(n-x) = 9!/2!(9-2)! multiply .15 2 (1-.15)(9-2) =.
Question 2
12 / 12 pts
You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) Find each of the following probabilities: a. Find P(Z ≤ .17). b. Find P(Z ≥ -.34). c. Find P(-1.14 ≤ Z ≤ 0.55). Your Answer: a)P(Z ≤ .17)=. b)P(Z ≥ -.34) =1.00-P(Z ≤ -.34)=1.00-.36693=. c) P(-1. ≤
a) Less than 4.5 inches? b) Greater than 4.0 inches? c) Between 3.8 inches and 4.7 inches? Your Answer: a) Less than 4.5" Z= x − μ σ = 4.5-4.4/.5 = .2 P(Z ≤ .2)=. b) Greater than 4.0" Z= x − μ σ = 4.0-4.4/.5=-.8 P(Z ≥ -.8) we look for P(Z ≤ -.8)= 1.0-.21186=. c) 3.8 and 4. Z= x −
μ σ = 3.8-4.4/.5=-1. Z= x − μ σ =4.7-4.4/.5=. P(-1. ≤ Z ≤ .6) P(Z ≤ -1.2)- P(Z ≤ .6) P(-1. ≤ Z ≤ .6)= .72575-.11507=. a. We have μ=4.4 and σ=.5.
P(-1.2 ≤ Z ≤ .6)=.72575-.11507=.61068.
Question 4
12 / 12 pts
You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) 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
Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data. Your Answer: E(x)= μ = Σ xf(x) 0(.18)+1(.10)+2(.38)+3(.22)+4(.07)+5(.05)= 2..
Expected # of Ins. policies that the salesp. will sell per day is 2. Variance (x)= σ = Σ ( x − μ ) 2 f(x) = (0-2.05 ) 2 (.18)+(1-2.05 ) 2 (.10)+(2-2.05 ) 2 (.38)+(3-2.05 ) 2 (.22)+(4-2.05 ) 2 (. )+(5-2.05 ) 2 (.05)= 1. Standard Dev: σ √ σ 2 =
= 1. The expected value is given by E(x)=μ=Σxf(x)= =0(.18)+1(.1)+2(.38)+3(.22)+4(.07)+5(.05)=2.05. So, the insurance salesperson can expect to sell 2. policies per day. The variance is given by
px(1-p)(n-x) =14!/7!(14-7)! multiply .55 7 (1-.55)(14-7) =.
