Answer:
(a) P(X=1) = 0.3079
(b) P(X≥1) = 0.3965
(c) P(X≥2) = 0.0886
(d) P(X≤1.9) = 0.9114
(e) Expected no. of hours = 3.594 hours
Step-by-step explanation:
We have,
p = 0.02
n = 25
q = 1-p
q = 0.98
We will use the binomial distribution formula to solve this question. The formula is:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure
Let X be the number of students who received a special accommodation.
(a) P(X=1) = ²⁵C₁ (0.02)¹ (0.98)²⁵⁻¹
= 25*0.02*0.61578
P(X=1) = 0.3079
(b) P(X≥1) = 1 - P(X<1)
= 1 - P(X=0)
= 1 - (²⁵C₀ (0.02)⁰ (0.98)²⁵⁻⁰)
= 1 - 0.6035
P(X≥1) = 0.3965
(c) P(X≥2) = 1 - P(X<2)
= 1 - [P(X=0) + P(X=1)]
= 1 - (0.6035 + 0.3079)
= 1 - 0.9114
P(X≥2) = 0.0886
(d) The probability that the number among 25 who received a special accommodation is within 2 standard deviations of the expected number of accommodations. This means we need to compute the probability P(X-μ≤2σ). For this we need to calculate the mean and standard deviation of this distribution.
μ = np = (25)*(0.02) = 0.5
σ = [tex]\sqrt{npq}[/tex] = √(25)*(0.02)*(0.98) = √0.49 = 0.7
P(X-μ≤2σ) = P(X - 0.5≤ 2(0.7)) = P(X≤ 1.4 + 0.5) = P(X≤1.9)
P(X≤1.9) = P(X=0) + P(X=1)
= 0.6035 + 0.3079
P(X≤1.9) = 0.9114
(e) Student who does not receive a special accommodation i.e. X=0 is given 3 hours for the exam whereas an accommodated student P(X>0) is given 4.5 hours. The expected average number of hours given on the exam can be calculated as:
Expected no. of hours = ∑x*P(x)
= 3*P(X=0) + 4.5*P(X>0)
= 3*0.6035 + 4.5(1 - P(X≤0))
= 1.8105 + 4.5(1 - 0.6035)
= 1.8105 + 1.78425
Expected no. of hours = 3.594 hours
Answer:
The solution the the given problem is given below.
Step-by-step explanation:
We have p = P(a student received a special accommodation) = 0.02,
so with X = the number among the 25 who received a special accommodation, X ∼ Bin(25, 0.02).
a) The probability that exactly 1 received a special accommodation is :
P(X = 1) =
[tex](0.02)^{1}[/tex] [tex](1-0.02)^{25-1}[/tex] = 25(0.02)(0.098)[tex]^{24}[/tex] ≈ 0.3079
b) The probability that at least 1 received a special accommodation is:
P(X ≥ 1) = 1 − P(X = 0) = 1 − (0.98)25 ≈ 1 − 0.6035 = 0.3965
c) The probability that at least 2 received a special accommodation is :
P(X ≥ 2) = 1 − P(X = 0) − P(X = 1) ≈ 1 − 0.6035 − 0.3079 = 0.0886
d) The mean and the standard deviation of X are:
µ = E(X) = np = 25(0.02) = 0.5,
σ = [tex]\sqrt{V(X)}[/tex] = [tex]\sqrt{np(1-p)}[/tex] = [tex]\sqrt{25(0.02)(0.98)}[/tex] = [tex]\sqrt{0.49}[/tex] = 0.7
Then, the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the expected number is :
P(|X − µ| ≤ 2σ) = P(X ≤ µ + 2σ) = P(X ≤ 0.5 + (2)(0.7)) = P(X ≤ 1.9)
= P(X = 0) + P(X = 1) ≈ 0.6035 + 0.3079 = 0.9114.
e) The average time allowed the 25 selected students will be:
[tex]\frac{(.5*4.5)+(24.5*3)}{25}[/tex] = 3.03 hours.
