Question

Need 9-12, 19 answered please!!

(9-12) Thirty-seven (37%) percent of students who major in Kinesiology do not work to subsidize their education whereas 45% of these students work at Bill Miller making $7.75 an hour and the remaining students work at West Telemarketing making $11.25 an hour. Let X be the hourly income of Kinesiology majors with the following probability distribution given by p(x) What is the mean hourly wage for this population of Kinesiology students? (rounded) o. What is the mode of hourly wages for this population of Kinesiology students? (rounded) I1. What is the population standard deviation of the hourly wages of Kinesiology students? 12. What is the probability that a randomly chosern Kinesiology major makes more than $10 given that he or she makes at least minimum wage (S6.25/hr.)?

Answer 1

Ans 9) Mean hourly wage for the students is given by

Mean = $\small \sum x_i * p(x_i)$ where x_i denote the different classes of hourly wage and p(x_i ) is the corresponding probability for it.

Therefore mean = 0 * 0.37 + 7.75*0.45 + 11.25*0.18

= 5.5125 dollars

Ans 10 ) The mode for such a discrete distribution is the value which is most frequent since p(x_i ) is highest for 7.75 dollars that is 0.45 so the mode of hourly wages for students is 7.75 dollars.

Ans 11) Population variation Var(X) is given by the formulae

$\small E(X^2) - E(X)^2$ where E(X) is the mean calculated and

$\small E(X^2) = \sum x_i^2 * p(x_i)$   

Therefore E(X²) = 0 * 0.37 + 7.75² * 0.45 + 11.25² * 0.18

= 49.809375

Therefore Var(X) = 49.809375 - 5.5125²

= 19.42171875

Standard Deviation for the Population is given by SD(X) = Var(X)^0.5 = 19.42171875^0.5 = 4.4070

Ans 12 Here we need to calculate the probability that a random chosen student makes more than 10 dollars that is X > 10 given that they earn atleast 6.25 dollars X>= 6.25 that is

P( X>10 | X>=6.25)

And the formula for Conditional Probability is P(X|Y) = P(X and Y) / P(Y) for two events X and Y

Therefore P(X>10 | X >= 6.25 ) = P(X> 10 and X >= 6.25) / P(X>6.25) = P(X>10) / P(X>=6.25)

Now P(X>10) = P(X=11.25) = 0.18 and P(X>=6.25) = P(X=7.75) + P(X=11.25) = 0.63

P( X>10 | X>=6.25) = .18/.63 = 0.2857

Ans 19) Let the event of KH finding broken spindle be denote by S and the event that spindle is supplied by A , B or C vendor be denoted by A , B or C respectively

So we need to Calculate P( A | S ) that is given the spindle is broken the probability that the spindle is provided by A

From the formula of Conditional Probability we can say

P(A | S) * P(S) = P ( A and S) = P( S|A ) * P(A)

Therefore P(A|S) = P(S|A) * P(A) / P(S)

P(S|A) = 0.01 and P(A) = 0.6

And from Bayes's Theorem P(S) = P(S|A) * P(A) + P(S|B) * P(B) + P(S|C) * P(C)

= 0.01 * 0.6 + 0.02 * 0.3 + 0.03 * 0.1 = 0.015

Therefore P(A|S) = P(S|A) *P(A) / P(S) = 0.01 * 0.6 / 0.15 = 6/15 = 0.4

Therefore the required probability is 0.4

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