Question

You manage a training program that has a failure rate of 34%. A new class of 6 students just checked in at the end of the fiscal year and you need to graduate at least 3 students. Your boss wants to know how likely it is that you will meet your yearly quota.

	x	P(x)	Expected Value	Std. Dev.
A	0
B	1
C	2
D	3
E	4
F	5
G	6

Calculate the expected number of students that will pass.Show your work. You can do so via a table or calculation.

Calculate the standard deviation. Show your work. You can do so via a table or calculation.

What is the probability of 4 students passing the training program? Write the calculator function and numbers that you used in your work.

What is the probability that at least half the class will pass? Write the calculator function and numbers that you used to show your work. T/F

Please show your work

Answer 1

Please don't hesitate to give a "thumbs up" for the answer in case the answer has helped you

p(pass) = 1-P(failure) = 1-.34 = .66

The event follows binomial distribution, B(n,p) = B(6, .66)

Also, number of trials = 6, So, n =6

Table for your reference:

X	p(x) = 'BINOM.DIST(x, 6, .66, FALSE)	Mean	Stdev
0	0.0015	3.96	1.1603
1	0.0180
2	0.0873
3	0.2260
4	0.3290
5	0.2555
6	0.0827

1.

Expected number of students that will pass: n*p = 6*.66 = 3.96

2.

Standard deviation = sqrt(n*p*(1-p)) = sqrt(6*.66*.34) = 1.1603

3.

P(4 pass ) = ?

Calculator pdf function: binompdf(n, p, x), where n = 6, p = .66, x = 4.

So, P(4 pass) = 6C4*.66^4 *34^2 = 0.3290

Answer: 0.3290

4.
Atleast half is >=3, so we should calculate P(X>=3) = ?

P(X>=3)
= 1- P(X<3)
= 1- P(X<=2)

The calculator function we use the binomcdf function binomcdf(n, p, x) = binomcdf(6, .66, 2)
We n = 6, p= .66, x = 2

= 1- 0.1069
= 0.8931

Answer: 0.8931