ISEN 350 Problem Set M6
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Texas A&M University *
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Course
350
Subject
Statistics
Date
Feb 20, 2024
Type
Pages
5
Uploaded by DoctorRook5940
Problem 1:
E(cx) = cE(x)
: This identity is related to the linearity of expectation. It states that the expected
value of a constant times a random variable is equal to the constant times the expected value of
the random variable. Mathematically:
E(cx) = ∫(cx) * f(x)dx where f(x) is the probability density function of x
= c ∫x * f(x) dx = cE(x) So, the identity is verified.
E(x + c) = E(x) + c:
This identity states that the expected value of the sum of a random variable
and a constant is equal to the expected value of the random variable plus the constant.
Mathematically:
E(x + c) = ∫(x + c) * f(x) dx
= ∫x * f(x) dx + ∫c * f(x) dx
= E(x) + c ∫f(x) dx = E(x) + c * 1 since ∫f(x) dx = 1 for a valid probability density function
= E(x) + c So, the identity is verified.
V(x) = E(x^2) - [E(x)]^2:
This identity represents the variance of a random variable. It states
that the variance of x is equal to the expected value of x squared minus the square of the
expected value of x. Mathematically:
V(x) = E[(x - E(x))^2]
= E[x^2 - 2x*E(x) + (E(x))^2]
= E(x^2) - 2E(x*E(x)) + E((E(x))^2)
E(xE(x)) = E(x) * E(E(x)) Since E(E(x)) is just E(x) again (the expected value of a constant is the
constant itself), we can simplify this further:
E(x*E(x)) = E(x) * E(x) = (E(x))^2 Replace E(x*E(x)) with (E(x))^2 in the expression:
V(x) = E(x^2) - 2(E(x))^2 + E((E(x))^2) Now, consider the last term, E((E(x))^2). Since E(x) is a
constant with respect to the expectation operator E, we have:
E((E(x))^2) = (E(x))^2 Replace E((E(x))^2) with (E(x))^2 in the expression:
V(x) = E(x^2) - 2(E(x))^2 + (E(x))^2
Simplify the expression: V(x) = E(x^2) - (E(x))^2
V(c + x) = V(x):
This identity states that adding a constant to a random variable does not
change its variance. Mathematically:
V(c + x) = E[(c + x)^2] - [E(c + x)]^2
= E[(c^2 + 2cx + x^2)] - [(c + E(x))^2]
= E(x^2) - [E(x)]^2 = V(x) So, the identity is verified.
V(cx) = c^2 * V(x)
: This identity relates to the variance of a constant times a random variable.
Mathematically:
V(cx) = E[(cx)^2] - [E(cx)]^2
= E(c^2 * x^2) - [c * E(x)]^2 = c^2 * E(x^2) - c^2 * [E(x)]^2
= c^2 * [E(x^2) - [E(x)]^2] = c^2 * V(x) So, the identity is verified.
Problem 2:
E[g(X)] = E[2X^2 − 2Xμ]
E[g(X)] = 2E[X^2] − 2μE[X]
E[X^2] = σ^2 + μ^2 and E[X] = μ:
E[g(X)] = 2σ^2 + 2μ^2 − 2μ^2 = 2σ^2
Therefore E[g(X)] is twice the variance of X
Problem 3:
(a) What is the probability that a sample of 10 covers will contain exactly 2 defectives?
𝑃(?
=
2)
=
(10 ?ℎ???? 2)
* (0. 10)
2
* (1 − 0. 10)
8
= 0. 1937
So, the probability that a sample of 10 covers will contain exactly 2 defects is approximately
0.1937.
(b) What is the probability that a sample of 10 covers will contain more than 2 defectives?
𝑃(?
=
1)
=
(10 ?ℎ???? 1)
* (0. 10)
1
* (1 − 0. 10)
9
= 0. 387420489
𝑃(?
=
0)
=
(10 ?ℎ???? 0)
* (0. 10)
0
* (1 − 0. 10)
10
= 0. 3486784401
𝑃(? > 2) = 1 − 0. 1937 − 0. 387420489 − 0. 3486784401 = 0. 07 (c) What is the probability that the first defective cover will be found on the 2nd cover
produced, given that all covers are being inspected?
𝑃(? = 2) = (1 − 0. 10)
2−1
* 0. 10 = 0. 09
(d) What is the probability the 4th part cover selected will be defective, given that the 3rd
cover selected was defective?
𝑃(4?ℎ ??????𝑖𝑣?|3?? ??????𝑖𝑣?) = 0. 10
(e) What is the probability the 2nd defective cover will be found on the 5th cover
sampled?
𝑃(2?? ????𝑙 ??????𝑖𝑣?? ?? 5?ℎ) = (5 ?ℎ???? 1) * (0. 10) * (0. 9)
4
* 0. 10 = 0. 03281
Problem 4
(a) A yard of cloth contains 2 or more blemishes.
𝑃(? ≥ 2) = 1 − 𝑃(? = 0) − 𝑃(? = 1)
𝑃(? ≥ 2) = 1 − (?
−0.1
0. 1
0
/0!) − (?
−0.1
0. 1
1
/1!) = 0. 00467
(b) A bolt contains less than 3 blemishes. Blemishes per bolt = 0.1 * 25 = 2.5
𝑃(? < 3) = (?
−2.5
2. 5
0
/0!) + (?
−2.5
2. 5
1
/1!) + (?
−2.5
2. 5
2
/2!) = 0. 54381
(c) A batch of 10 bolts contains more than 25 blemishes. Blemishes per 10 bolts = 25
𝑃(? > 25) = 1 −
𝑖=0
25
∑ (?
−25
25
𝑖
/𝑖!)
=
1
−
0. 55292
=
0. 44708
Problem 5
(a) E(2X - 0.002):
E(2X - 0.002) = 2E(X) - E(0.002)
E(2X - 0.002) = 2(0.024) - 0.002 = 0.048 - 0.002 = 0.046
So, E(2X - 0.002) = 0.046.
(b) V(3X - 0.01):
V(3X - 0.01) = (3^2) * V(X) V(X) is the variance of X, which is 0.0045:
V(3X - 0.01) = (3^2) * 0.0045 = 9 * 0.0045 = 0.0405
So, V(3X - 0.01) = 0.0405.
(c) E(2X^2):
E(2X^2) = 2E(X^2)
E(2X^2) = 2[V(X) + E(X)^2]
E(2X^2) = 2(0.0045 + 0.024^2) = 0.010
E(2X^2) = 0.010
3.23. A random sample of 50 units is drawn from a production process every half hour.
The fraction of nonconforming products manufactured is 0.02. What is the probability
that p̂ ≤ 0.04 if the fraction nonconforming really is 0.02?
𝑃(?̂ ≤ 0. 04) = 𝑃(𝑥 ≤ 2) =
𝑖=0
2
∑ ((50 ?ℎ???? 𝑖)
* (0. 02)
𝑖
* (0. 98)
50−𝑖
) = 0. 9216
So, the probability that p̂ ≤ 0.04 is 92.16%
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