In a particular faculty 60% of students are men and 40% are women. In a random sample of 50 students what is the probability that more than half are women?Let the random variable X = number of women in the sample. Assume X has the binomial distribution with n = 50 and p = 0.4.a. What is the expected value and variance of the random variable X?b. Calculate the exact binomial probability. c. Are the conditions that permit you to use a normal approximation to the binomial satisfied? Explaind. Recalculate the probability in part b using a normal approximation without the continuity correction. e. Recalculate the probability in part b using a normal approximation with the continuity correction.

a) The expected value is given by:

and the variance is given by:

b)

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

c) 1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

d)

e)

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

The probability mass function for the Binomial distribution is given as:

Where (nCx) means combinatory and it's given by this formula:

Part a

The expected value is given by:

and the variance is given by:

Part b

For this case we want to find this probability:

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

Part c

1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

Part d

We want this probability:

Part e

For this case we use the continuity correction and we have this:

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation:

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632