Question

Stats ICA4

NAME:

A college professor tells his class what the probability of getting an A on any one of his tests/quizzes is, based on his 40 years of prior results. The probability is 85%. He gives his class 4 tests/quizzes per semester. What is the probability that a random student gets an A on 3 or more of those tests/quizzes? Please show all work.
The following procedures DO NOT result in a binomial distribution. For each case, explain why.
Jaimie decides to give 4 students in his class an A. In the interest of limiting gender bias, he decides to select 2 males and 2 females. As he selects each student individually (at random), one girl calculates the probability that she will be selected, i.e. she is making the calculation every time someone is picked (hint hint, wink wink, nudge nudge).
At a group therapy session, each of Peter Griffin’s three children are asked if they are happy or not with Peter’s parenting skills (a "yes" or "no" answer only).
Twenty random students are selected from a class of 50 students, one at a time. They are asked to rate the textbook on a scale of 1 to 10.
A professor has a 95% passing rate for all of his past classes. He surveys his current 93 students, and 13 of them are worried about failing.
For his class of 93 students, determine the mean and standard deviation of students passing the class.
Subtract: Mean – 1 Standard Deviation =
Add: Mean + 1 Standard Deviation =
What do the numbers from parts B & C actually represent in words? Based on those results, do you feel that 13 students worrying about failing is reasonable? Explain.

Answer 1

Answer:

1) Pr (more than 3) = 0.890

2) We can tell if a procedures result in a binomial distribution or not by checking if it satisfies these four binomial distribution conditions

3)

Step-by-step explanation:

n = 4

p= 85% = 0.85

nCx = n Combination x

Probability = nCx* p^x *(1-p)^(n-x)

Pr (more than 3) = Pr(3) + Pr(4)

Pr(of having A) = 85% = 0.85

x = 3

Pr(having A in 3quizes) = 4C3 * 0.85^3 * (1-0.85)^(4-3)

= 4 * 0.614125 * 0.15

= 0.368

x = 4

Pr(having A in 4quizes) = 4C4 * 0.85^4 * (1-0.85)^(4-4)

= 1 * 0.522 * 0.15^0 = 1 * 0.5522 * 1

= 0.522

Pr (more than 3) = 0.368 +0.522

Pr (more than 3) = 0.890

We can tell if a procedures result in a binomial distribution or not by checking if it satisfies these four binomial distribution conditions:

1) A fixed number of trials

2) Each trial is independent of the others

3) There are only two outcomes

4) The probability of each outcome remains constant from trial to trial.

First case does have independence. The probability is calculated based on the outcome of other students. That is the probability of picking the girl is affected by other students picked before her.

Second case: it doesn't have a fixed number of trials

Third case: The experiment outcome is more than two. The experiment doesn't have a fixed number of independent trials.

Professor:

p = 95% = 0.95

n = 93 students

13 students are worried of failing = 5% of 93 students = 4.65

Mean = np = 0.95 ×93

Mean = 88.35

Standard deviation = √[np(1-p)]

= √(88.35×(1-0.95))

=√(88.35×0.05) = √83.9325

Standard deviation = 9.16

Mean – 1 Standard Deviation = 88.35-9.16 = 79.19

Mean + 1 Standard Deviation = 88.35+9.16 = 97.51

The standard deviation of datasets which have normal distribution can be used to find the proportion of values that lie within a particular range of the mean value.

68% of the values in the dataset will lie between Mean – 1 Standard Deviation and Mean + 1 Standard Deviation