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
