Question

Question 3

1. A merchandizing person at Game Store warehouse opens a box with 250 parts contained in it (see table above). What is the probability that two (2) randomly selected parts will be from a supplier X and supplier Y? Assume part replacement method is applied.                                                                                                                                                         (3)
2. Assume the part was not replaced in the box before the second part was selected. What is the probability?                                                                                                             (3)
3. If the probability of obtaining a three (3) on a six-sided die is 1/6 = 0.167, compute the probability of obtaining any number but a three (3)                                                (3)
4. If the probability for X number of students to call the lecturer for QUT151Z on the day before the pre-lim exam, is given in the table below. Compute the expected number of students who will call the lecturer a day before the final exam.                           (3)
5. An experienced car sales person knows that he can sell two (2) cars to two customers out of ten (10) who entered the dealership. Determine the probability that he will sell a car to exactly two (2) of the next three (3) customers.                                                      (3)

    The mean value for the weight of a specific box of pasta for the past 6 months is 0.297 kg with a standard deviation of 0.024 kg. Assuming a normal distribution, find the percent of the data that falls below the lower specification limit of 0.274 kg as well as the percentage that falls above 0.347 kg.

Answer 1

(Since there are more than 4 parts i will answer first 4)

i.

let us assume no. of parts from supplier X = m

then , no. of parts from supplier Y = 250-m

event A : part form x and 1 part from y

P(A) = P(first part from X second part from Y) + P(first part from Y second part from X)

= P(X)*P(Y) + P(Y)*P(X)

= (m/250)*((250-m)/250) + ((250-m)/250)*(m/250)

P(A) = 2(m)*(250-m)/(250^2)

ii.

event A : part form x and 1 part from y

P(A) = ((choose 1 from X)*(choose 1 from Y)) / (chhose 2 from total)

= ((mC1)*((250-m)C1)) / (250C2)

= m*(250-m) / (250*249/2)

P(A) = 2(m)*(250-m)/(250*249)

iii.

P(X) = p

then P(not X) = 1-p

here,

P(3) = 0.167

therefore,

P(not 3) = 1-0.167 = 0.833

P(not 3) = 0.833

iv.

E = sum of (x.P(x))

E(no. of students call lecturer a day before exam) =sum of ( x.P(x students call lecturer a day before exam))

data of x and P(x) not present so leave it in generalized form

E(no. of students call lecturer a day before exam) =sum of ( x.P(x students call lecturer a day before exam))

P.S. (please upvote if you find the answer satisfactory)