ACTIVITY SET 9 Stat 200 Sections 4-6 Feb. 16Activity 9.1 Work with a partner. Use one of the sets of four cards distributed during class. Verify that there is one card from each suit (hearts, diamonds, clubs, spades) in the set. Randomly mix the cards and then select one in a way that your partner can’t see what you picked. Have the partner guess the suit of thecard you picked. Repeat this 10 times, and keep track of right and wrong guesses using the following table. Then change places so that you do the guessing and your partner does the card mixing.Trial 1 2 3 4 5 6 7 8 9 10Right or Wrong?a. What was the number of correct guesses made by your partner? If you weren’t present for the lab, write“Not present” for your answer (we won’t hold it against you.)Number of correct guesses by partner = _____ If you did this activity during class, please enter the number of correct guesses by your partner into ANGEL using the specified link in the Week 6 folder. b. Assuming a person randomly guesses each time, explain why this is a binomial experiment and give the values of n and p for this experiment. c. Assuming random guessing, what is the “expected value” (mean value) of the number of correct guesses in this experiment? Show the calculation. d. Start Minitab (from Spreadsheets and Statistics folder of Start menu). Use Calc>Probability Distributions>Binomial. At the top of the resulting dialog box, click on “Probability,” then enter 10 as number of trials and 0.25 as the probability of success. Near the bottom, click on “Input Constant and enter 0 in the adjacent box. This will give the probability of 0 correct guesses after OK is clicked.Probability of 0 correct guesses = P(X = 0) = e. Use Minitab to find (separately), the probabilities of 1, 2, and 3 correct guesses.Probability of 1 correct guesses = P(X = 1) = Probability of 2 correct guesses = P(X = 2) = f. Add the probabilities found for 0, 1, and 2 to find the cumulative probability P(X ≤ 2), the probability that the number of correct guesses is 2 or less. (You can round off the probabilities to 3 decimal places if your wish.)Probability of 2 or less correct guesses = P(X ≤ 2) = g. Again use Calc>Probability Distributions>Binomial. NEAR THE TOP, click on “Cumulative Probability” and near the bottom change the Input Constant to 6. This will give the cumulative probabilityfor 6 or less correct guesses. Probability of 6 or less correct guesses = P(X ≤ 6) = h. Using the probability found in part g, determine the probability somebody could get more than 6 correct guesses.Activity 9.2 In each part, explain (1) whether the variable is discrete or continuous AND (2) if it is discrete whether it is a binomial random variable OR if it is continuous, whether you think it might be a normal random variable. Note – the issue of whether it’s normal will mainly be a “guess” on your part. Just explain why you think it might or might not be. a. Weights of all students at Penn State.b. Number of correct answers on a 30 question multiple choice test for somebody who randomly guesses at every question. There are four answer choices for each question,c. Heights of women at Penn Stated. A woman buys a lottery ticket every week for which the probability of winning anything at all is 1/10. She continues to buy them until she has won 3 times. X = the number of tickets she buys.e. Daily snowfall amounts in State College, for all days between December 1 and March 15. Activity 9.3 For a randomly selected sample of n = 10 people, assume X = number of people with type O+ blood is a binomial random variable with p =0.3 (the fraction with type O+ in the larger population). Use Minitab (as it was used in activity 9.2) to assist in answering the following.a. P(X = 5), the probability that the number of people with type O+ is exactly 5.b. P(X  5), the probability that the number with type O+ is 5 or less. Note that this is a cumulative probability.c. What is the probability that more than 5 people in the sample have type O+ blood? Activity 9.4 Suppose that n = 72 is the number of students in attendance for this section 6 today. Let’s assume that the chance any single person has their birthday today is p = 1/365 = .00274. If birthdays are independent, then X= number having a birthday today is a binomial random variable with n = 72 and p = .00274.Use Minitab to assist in answering the following.a. What is the probability that nobody among the n = 72 students has a birthday today? That is, find P(X=0) for a binomial random variable with n = 72 and p = .00274.b. What is the probability one or more students among the n = 72 students has a birthday