Question

Q1
In a class on 50 students, 35 students passed in all subjects, 5 failed in one subject, 4 failed in two subjects and 6 failed in three subjects.

Construct a probability distribution table for number of subjects a student from the given class has failed in.
Calculate the Standard Deviation.








Q2
45 % of the employees in a company take public transportation daily to go to work. For a random sample of 7 employees, what is the probability that at most 2 employees take public transportation to work daily?

















               
     Q3. Find
        a) P(z < 1.87)
        b) P(z > -1.01)
        c)   P(-1.01 < z < 1.87)
           






Q4
Assume the population of weights of men is normally distributed with a mean of 175 lb. and a standard deviation 30 lb. Find the probability that 20 randomly selected men will have a mean weight that is greater than 178 lb.



















   
Q5
We have a random sample of 100 students and 75 of these people have a weight less than 80 kg. Construct a 95% confidence interval for the population proportion of people who have a weight less than 80 kg.










Q6
We have a sample of size n = 20 with mean x ̅ =12 and the standard deviation σ=2. What is a 95% confidence interval based on this sample?

Answer 1

1) P(X = 0) = 35/50 = 0.7

P(X = 1) = 5/50 = 0.1

P(X = 2) = 4/50 = 0.08

P(X = 3) = 6/50 = 0.12

E(X) = 0 * 0.7 + 1 * 0.1 + 2 * 0.08 + 3 * 0.12 = 0.62

E(X^2) = 0^2 * 0.7 + 1^2 * 0.1 + 2^2 * 0.08 + 3^2 * 0.12 = 1.5

Variance = E(X^2) - (E(X))^2

               = 1.5 - (0.62)^2

               = 1.1156

Standard deviation = sqrt(1.1156) = 1.0562

2) p = 0.45

    n = 7

It is a binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

               = 7C0 * (0.45)^0 * (0.55)^7 + 7C1 * (0.45)^1 * (0.55)^6 + 7C2 * (0.45)^2 * (0.55)^5

               = 0.3164

3) a) P(Z < 1.87) = 0.9693

    b) P(Z > -1.01) = 1 - P(Z < -1.01)

                            = 1 - 0.1562

                            = 0.8438

c) P(-1.01 < Z < 1.87)

= P(Z < 1.87) - P(Z < -1.01)

= 0.9693 - 0.8438

= 0.1255

4) P( > 178)

= P(( - )/() > (178 - )/())

= P(Z > (178 - 175)/(30/))

= P(Z > 0.45)

= 1 - P(Z < 0.45)

= 1 - 0.6736

= 0.3264

5) = 75/100 = 0.75

At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval for population proportion is

+/- z_0.025 * sqrt((1 - )/n)

= 0.75 +/- 1.96 * sqrt(0.75 * 0.25/100)

= 0.75 +/- 0.085

= 0.665, 0.835

6) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval for population mean is

+/- z_0.025 *

= 12 +/- 1.96 * 2/

= 12 +/- 0.8765

= 11.1235, 12.8765