Need help in Mean, Probability & SD

[sumans]

The placement data of a business school suggest that the percentage of students securing a job with a multinational during the years 1996, 1997 and 1998 are 20%, 30% and 40% respectively. One student is chosen at random from the graduating batch of each year. Let X be the number of students, out of the three so selected who had secured a job with a multinational. Calculate the following:

a) The probability distribution of X
b) The Mean of X
c) The Standard Deviation of X

 

[panky05]

Can anyone post answer of this?

The placement data of a business school suggest that the percentage of students securing a job with a multinational during the years 1996, 1997 and 1998 are 20%, 30% and 40% respectively. One student is chosen at random from the graduating batch of each year. Let X be the number of students, out of the three so selected who had secured a job with a multinational. Calculate the following:

a) The probability distribution of X
b) The Mean of X
c) The Standard Deviation of X

 

[SW VandeCarr]

Can anyone post answer of this?

Are you asking for sumans? Can't sumans speak for him/herself? In any case, his/her question is over 4 years old and if this was a homework question, it should have been posted in that forum. When posting questions, the poster should show an attempt at a solution.

 

[micromass]

This thread is 4 years old.

 

[blue_raver22]

a) The probability distribution of X can be calculated by using the given percentages and the fact that one student is chosen at random from each year's graduating batch. The possible outcomes for X are 0, 1, 2, and 3, representing the number of students who secured a job with a multinational. The probabilities for each outcome can be calculated as follows:

P(X=0) = (1-0.2)(1-0.3)(1-0.4) = 0.42
P(X=1) = (1-0.2)(1-0.3)(0.4) + (1-0.2)(0.3)(1-0.4) + (0.2)(1-0.3)(1-0.4) = 0.42
P(X=2) = (1-0.2)(0.3)(0.4) + (0.2)(1-0.3)(0.4) + (0.2)(0.3)(1-0.4) = 0.24
P(X=3) = (0.2)(0.3)(0.4) = 0.024

Therefore, the probability distribution of X is:

X | 0 | 1 | 2 | 3
---|---|---|---|---
P(X) | 0.42 | 0.42 | 0.24 | 0.024

b) The mean of X can be calculated by multiplying each outcome by its corresponding probability and adding them together. This can also be represented as the expected value of X.

Mean of X = E(X) = (0)(0.42) + (1)(0.42) + (2)(0.24) + (3)(0.024) = 0.42 + 0.42 + 0.48 + 0.072 = 1.392

Therefore, the mean of X is 1.392.

c) The standard deviation of X can be calculated by first finding the variance, which is the average of the squared differences between each outcome and the mean. This can be represented as Var(X) = E[(X-E(X))^2]. The standard deviation is then the square root of the variance.

Var(X) = (0-1.392)^2(0.42)