I need help with this question 1. From past experience, a company has found that
ID: 3200230 • Letter: I
Question
I need help with this question
1. From past experience, a company has found that in carton of transistors, contain transistors, 92% no defective transistors, 3% contain one defective transistor, 3% contain two defective and 2% contain three defective transistors. a. Construct a probability distribution below. P(X) b. Calculate the mean, variance, and standard deviation for the defective transistors. 2. The number of suits sold per day at Suit world isshown in the probability distribution below 19 20 21 22 23 POx) 0.2 0.2 0,3 0.2 0.1 a. Find the mean, variance, and standard deviation of the distribution. b. If the manager of Suit World wants to make sure that he has enough suits for the next five days, how many should he buy to stock the store?
Explanation / Answer
Back-up Theory
1. Percentage composition is equivalent to probability.
2. Given the probability distibution, i.e., given {x, p(x)},
Mean = E(X) = Sum(over all possible values of x){x.p(x)}
Variance = V(X) = E(X2) - {E(X)}2 = Sum(over all possible values of x){x2.p(x)} - {E(X)}2
Now, for the solution,
Q1a
Probability Distribution
Let X = Number of defective transisters per carton and p(x) = percentage of cartons having x number of defectives per carton. Then, {x, p(x)} gives the probability distribution of number of defectives per carton
Q1(b) Calculation of Mean and Standard Deviation
Mean = µ = E(X) = 0.15 [from the above table] ANSWER
Variance = 2 = E(X2) - {E(X)}2 = 0.33 - 0.152[from the above table]
= 0.3075
Standard Deviation = = sq.rt(0.3075) = 0.5546 ANSWER
Q2(a)
Mean = µ = E(X) = 20.8 [from the above table] ANSWER
Variance = 2 = E(X2) - {E(X)}2 = 434.2 - 20.82[from the above table]
= 1.56
Standard Deviation = = sq.rt(1.56) = 1.25 ANSWER
Q2(b)
Required Stock = 5 x Average Sales per day = 5 x 20.8 = 104 ANSWER
Probability Distribution x 0 1 2 3 Total p(x) 0.92 0.03 0.03 0.02 1.00