Solve the following questions 6×3 mark=18 mark 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. Solution: 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? Solution: 3. Find a) P(z -1.01) c) P(-1.01 < z < 1.87) Solution: 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. Solution: 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. Solution: 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? Solution:

Solve the following questions                                                  6×3 mark=18 mark

1. 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.

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Solve the following questions 6×3 mark=18 mark 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. Solution: 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? Solution: 3. Find a) P(z -1.01) c) P(-1.01 < z < 1.87) Solution: 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. Solution: 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. Solution: 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? Solution:
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1. Construct a probability distribution table for number of subjects a student from the given class has failed in.
2. Calculate the Standard Deviation.

Solution:

 No of subjects failed(x) Probability 1 0.1 2 0.08 3 0.12 Total 0.3

Variance=x2px

= (1×0.1) + (4×0.08) + (9×0.12)

=1.5

Standard deviation= sqrt (Variance)

Standard deviation=1.2247

1. 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?

Solution:

This is a Binomial trial. The probability of success, p=0.45, q=0.55 and n=7

px≤2=px=0or px=1orp(x=2)

px≤2=∑nxpxqn-x

px≤2=700.4500.557+710.4510.556+720.4520.555

=0.015224+0.087194+0.214022

=0.31644

1. Find
2. a) P(z < 1.87)
3.        b)  P(z > -1.01)
4.        c) P(-1.01 < z < 1.87)

Solution:

• Pz < 1.87=.9693
• Pz > -1.01=1-p(z<-1.01)

=1- 0.1562

=0.8438

• =0.9693-0.8438

=0.1255

1. 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.

Solution:

px>178=1-p(x-μsn)

px>178=1-p(178-1753020)

=1-P(0.4472)

=1-p (0.45)

=1-0.6736

=0.3264

1. 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.

Solution:

P=75/100

=0.75

Z=1.96

Confidence interval=pz∝/2p(1-p)n

Confidence interval=0.75∓1.960.75(1-0.75)100

=0.75∓0.0849

=(0.6651, 0.8349)

1. 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?

Solution:

Confidence interal=xt∝/2sn

Confidence interal=12∓2.093220

=12∓0.936

=(11.064, 12.936)