Statistics Worksheet Coursework Example

INSTRUCTIONS:
This is not to be written as a paper. Please insert the answers to the questions on this document.
Showing work is required! If tables are required insert them.
If Excel is used please attach the work book and label each tab with the problem and question it represents. Ensure all data is there including formulas.
If a sketch is required please ensure it is scanned and labeled accordingly. Thank you!
WK3 WS1 Discrete Probability Distributions
Read the following scenario and complete each of the four problem sets below:
Suppose the distribution below represents the probability of a person to make a certain grade in Chemistry 101, where x = the letter grade of the student in the class (A=4, B=3, C=2, 0=1, F=O).

Is this an example of a valid probability distribution? Explain your answer. This is a valid probability distribution because probabilities add up to 1.
Determine the probability that a student selected at random would make a C or higher in Chemistry P= 0.34*0.21*0.15 = 0.011
Determine the mean and standard deviation of the grade distribution. Mean = (0.08+0.22+0.34+0.21+0.15)/5 = 0.2
Standard deviation = 0.0963
Make a NEW table that shows only the probability of students passing or failing (i.e. receiving an F) Chemistry 101.

“A professor has 30 students in his or her Chemistry 101 course and wants to determine the probability that a certain number of these 30 students will pass his or her course.” Does this represent a binomial experiment? Explain your answer.

This is a binomial experiment because binomial experiment has two outcomes being expected which in this case is either fail or pass. Additionally, it has a fixed number of trials which is one.
Based on the Pass/Fail table, what is the probability that 24 of the 30 students enrolled in Chemistry 101 during the spring semester will pass the course? P(x) = 24/30 = 0.8
Determine the mean and standard deviation of the data in the Pass/Fail table.
WK3 WS2 Nominal Probability
Read the following scenario and complete each of the four problem sets below:
A.Use the z-table to determine the following probabilities. Sketch a normal curve for each problem with the appropriate probability area shaded.
1.P(z > 2.34) = 0.9904
2.P(z < -1.56) = 0.594
3.P(z = 1.23) = 0.8907
4.P(-1.82 < z < 0.79) = P(z ˂ 0.79) – P(z˂-1.82) = 0.7852 – 0.4286 = 0.3566
5.Determine the z-score that corresponds with a 67% probability.
=0.7486
B. Read the following scenario:
Park Rangers in a Yellowstone National Park have determined that fawns less than 6 months old have a body weight that is approximately normally distributed with a meanµ = 26.1 kg and standard deviation a = 4.2 kg. Let x be the weight of a fawn in kilograms.
Complete each of the following steps for the word problems below:
•Rewrite each of the following word problems into a probability expression, such as P(x>30). = P(x = 26.1)
•Convert each of the probability expressions involving x into probability expressions involving z, using the information from the scenario.
Z = x -µ / σ = z = (x – 26.1)/4.2
•Sketch a normal curve for each z probability expression with the appropriate probability area shaded.
•Solve the problem.
1. What is the probability of selecting a fawn less than 6 months old in Yellowstone that weighs less than 25 kilograms?
Z = 24 – 26.1/4.2 =-0.5. From z-score table, -0.5 is 0.3082
Probability of selecting fawn with weight less than 25kgs is 0.3082
2. What is the probability of selecting a fawn less than 6 months old in Yellowstone that weighs more than 19 kilograms?
Z = x – µ / σ = 19 – 26.1/ 4.2 = -1.69, probability is 0.0455. 3. What is the probability of selecting a fawn less than 6 months old in Yellowstone that weighs between 30 and 38 kilograms?
Z = (30 – 26.1)/ 4.2 = 0.93, p is 0.8238
Z = (38 – 26.1) / 4.2 = 2.83 , p is 0.9977
P = 0.9977 – 0.8238 = 0.1739
4. If a fawn less than 6 months old weighs 16 pounds, would you say that it is an unusually small animal? Explain and verify your answer mathematically.
Z = (16 – 26.2) / 4.2 = -2.43, p = 0.0075. This is a small animal because probability of selecting it is low.
5. What is the weight of a fawn less than 6 months old that corresponds with a 20% probability of being randomly selected? Explain and verify your answer mathematically.
P = 0.2 (20%) then z score from the table is -0.84
-0.84 = (x – µ) / 4.2
= 4.2 * -0.84 = x – µ where µ is 26.1
X = 22.60kgs