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STT351 Final Exam
Name: PID:
- The probability mass function p(x) of a discrete random variable X is p(0) = .15, p(1) = .30, p(2) = .20, p(3) = .10, and p(4) = .25, then the value of the cumulative distribution function F (x) at x = 2 is.
Answer: F (2) = P (X ≥ 2) = P (0) + P (1) + P (2) = .15 + .3 +. 2.
- If the probability density function of a continuous random variable X is
f (x) =
9 x^0 ≤^ x^ ≤^3 0 otherwise
then, P (2 ≤ x ≤ 2 .5) is.
Answer:
2 29 xdx^ =^29 (^2.^5
(^2) − 22 2 ) =^.^25
- Let X be a exponential random variable with λ = 2. Then V (. 5 X+1.23) is.
Answer: V (X) = 1/λ^2 = 1/ 4. Then V (. 5 X + 1.23) = (.5)^2 V (X) = (.25)(1/4) = 1/ 16.
- Numerical value of z. 15 is.
Answer: 1.
- Suppose that A and B are independent events with P (A) = .40 and P (B) = .70, then P (B′|A′) is.
Answer: P (B′|A′) = P (A′^ ∩ B′)/P (A′) = (P (A′)P (B′))/P (A′) = P (A′) =. 3
- If P (A) = .45, P (B) = .60, and P (A∩B′) = .10, then P (A ∪ B) =.
Answer: Note that P (A ∩ B) = P (A) − P (A ∩ B′) =. 45 − .1 =. 35. Then, P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = .45 +. 6 − .35 =. 7.
- There are 20 laptops in identical boxes. 3 of them are known to be broken. If you randomly sold 5 laptops, probability that exactly 1 out of 5 sold is broken is .
Answer: (
(^205 )
- If P (A ∩ B) = .23 and P (A|B) = .60, then P (B) is.
Answer: Note that P (A ∩ B) = P (A|B)P (B). So P (B) =. 23 /.6 =. 383.
- Let X 1 ,... , X 100 be random sample Chi-squared distribution with 4 degrees of freedom. Then probability that is X will be less than 4 is.
Answer: P (1. 7 < Z < 2 .3) = Φ(2.22) − Φ(− 1 .61) = 0. 987 − 0 .054 = 0. 933.
- If the probability density function of a continuous random variable X is
f (x) =
ke−x^0 ≤ x < ∞ 0 otherwise
then the value of k is.
Answer:
0 ke−xdx^ = 1. That means^
ke−x − 1
∞ 0 = 1. So^ k^ = 1.
- If X 1 ,... , X 20 are random sample from the distribution Gam(α, β), with α known to be 2, Method of Moment Estimator for β is.
Answer: E(X) = αβ = 2β. Thus β = E(X)/ 2. That means βˆ = X/¯ 2.
- Suppose that you are playing a game that if a die you throw shows number bigger than 4, then you win $1. Otherwise you lose $1. You have decided to play this game 50 times. The probability that you would make a profit after the 50 games is .
Answer: P ( make profit) = P ( X >¯ 0). By CLT, we have X¯ ∼ N (μ, σ^2 /n), where μ = E(X) = 1(2/6) + (−1)(4/6) = − 1 / 3 , E(X^2 ) = 1^2 (2/6) + (− 12 )(4/6) = 1, and V (X) = E(X^2 ) − [E(X)]^2 = 1 − 1 / 32 = .889 Therefore, we have X¯ ∼ N (−. 333 ,. 889 /50). Now P ( X >¯ 0) = 1 − Φ{(0 − (−.333))(
- Let X 1 ,... , Xn be random sample from the normal distribution with variance σ^2 = 5. If we wish to estimate μ with an accuracy of ±.1 with 95% confidence, calculate the value of n needed.
Answer: We need (z. 025 σ/
n) =. 1 , when σ =
- That means n = 19.208. So we need n 20.
- If the distribution of a random variable is approximately normal, then about % of the values are within three standard deviation of the mean.
Answer: By 68-95-99.7 rule, it is 99.7%.
- Let X 1 ,... , X 250 be random sample from unknown distribution with variance σ^2 = 3. Sample mean was 2.3. Obtain lower-bound 95% (one-sided) confidence interval for μ.
Answer: One-sided CI is X¯ − z. 05 σ/√n = 2. 3 − (1.65)
250 = 2.12. So the answer is (2. 12 , ∞).
- Let X 1 ,... , X 15 be random sample from the normal distribution with mean μ and standard deviation σ. Sample mean was 1.3, and sample standard deviation was 3.14. Obtain 99% (two-sided) confidence interval for σ^2.
Answer: Two -sided CI is X¯ ± z. 005 σ/
n = 1. 3 ± (2.58)3. 14 /
15 =. So the answer is (− 0. 792 , 3 .392).
- Let X be a r.v. with joint pdf
f (x, y) =
15 e
−xy (^2) if 0 ≤ x, 1 ≤ y ≤ 4 0 otherwise
Then E(XY ) is.
Answer: E(XY ) =
0
1 xy^152 e−xy^2 dydx^ =^152
0
1 xe−xy^3 dydx. That means^ E(XY^ ) = 2 15
0
1 xe
−xy (^3) dydx = 2 15
0
1 xe
−x((4 (^4) − 14 )/4)dx = 2 15
44 − 1 4 (−e
- There are 7 red balls and 4 white balls in a box. 3 balls are chosen at once ( without order ). Number of ways of drawing [all 3 balls are of the same color] are.
Answer: There are
3
ways of drawing RRR, and
3
ways of drawing WWW. Add the two.
- Suppose that waiting time at local bank has Exponential distribution with λ = 1/5. The probability, that your waiting time will exceed 10 min is.
Answer: We have CDF of exponential distribution, P (X ≥ x) = 1 − e−λx. Using this, we have P (X > 10) = 1 − P (X ≥ 10) = e−^10 /^5 = e−^2.
- Suppose there are 9 players in the baseball team. Number of ways you can order them is.
Answer: 9! ways.
- Suppose you are trying to estimate proportion p, with accuracy of ±1%. With 95% confidence, how many sample size do you need?
Answer: Margin of error for p is ±zα/ 2
(ˆp(1 − pˆ))/n. Since we don’t know value of p or ˆp in this case, use 1/2 for ˆp, which is the case that margin of error is maximum. Then margin of error = ±(1.96)
(1/4)/n ≈ ± 1 /√n. Therefore we need n = 10, 000.
- Suppose that lifetime of a electrical parts is modeled by a exponential random variable with λ = .4 (in years). What is the probability that it will last more than half a year, but break down before 2 years?
Answer: P (. 5 < X < 2) = P (X < 2) − P (X < .5). We have CDF of exponential r.v. as P (X < x) = 1 − e−λx. So the answer is −e−.4(2)^ + e−.4(.5)^ =. 3694.
- Suppose that lifetime of a electrical parts, X, is a exponential random variable with λ = .4 (in years). If you have 5 identical parts, and assume that they are independet of each other, what is the probability that all of them will last more than 4 years?
Answer: Say for parts #1, P (X 1 > 4) = 1 − P (X 1 < 4) = e−.4(4). This is same for all 5 parts. Then
P ( all of 5 parts will last more than 4 yrs) = P ([X 1 > 4] ∩ [X 2 > 4] ∩ [X 3 > 4] ∩ [X 4 > 4] ∩ [X 5 > 4]) = P (X 1 > 4)P (X 2 > 4)P (X 3 > 4)P (X 4 > 4)P (X 5 > 4) = P ([X 1 > 4])^5 = e−.4(4)5^ = 0. 00034.
- If X 1 ,... , X 30 is a Normally distributed random variable with mean 4.4 and SD 1.5, what is probability that X will be in between 4.2 and 4.7?
Answer: Since X¯ ∼ N (4. 4 , 1. 52 ), we have P (4. 2 < X <¯ 4 .7) = Φ((4. 7 − 4 .4)/ 1 .5) − Φ((4. 2 − 4 .4)/ 1 .5) = 0. 579 − 0 .447 = 0. 132
- If Z is a standard normal random variable what is 10th percentile?
Answer: Since Φ(− 1 .28) = .1, 10th percentile of Z is -1.28.
- If X is a Normally distributed random variable with mean μ and SD 2, for what what value of μ, probability that X will be greater than 4.9 will be 80%?
Answer: We need P (X > 4 .9) = 1 − Φ((4. 9 − μ)/2) = .8. That means Φ((4. 9 − μ)/2) =
. 2. Since Φ(− 0 .842) = .2, equating (4. 9 − μ)/2 = − 0 .842, μ = 6. 584.
- There are four cards marked { 2 , 3 , 4 , 5 } in a box. Two cards are randomly drawn from the box at the same time. Let random variable X to be the DIFFERENCE between two numbers (larger number - smaller number). Let F (x) be CDF of random variable X. Then F (2) =. Answer: Since we are picking 2 cards from 4 available, we have
2
= 6 combinations in total. They are [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 5] Their difference will be (respectively) [1], [2], [3], [1], [2], [1] Since each of them has equal chance of being drawn, F (2) = P (X ≥ 2) = 5/ 6.
- A machine at AMT & Co. fills 120-ounce jugs with laundry softener. A quality control inspector wishes to know if the machine needs an adjustment or not. The machine needs adjustment if it’s overfilling the jugs on average. Assume the distribution of the amount of laundry softener in these jugs is Normal with mean μ and standard deviation of 1.5 ounces. The inspector is thinking of taking the random sample of n jugs, but have not decided on what value of n to use.
(a) If you want to estimate μ to the accuracy of ±.5(ounce) with 95% confidence level, how many test runs are needed?
Answer: Margin of error for μ is ±z. 025 σ/
n. So we need (1.96)1. 5 /
n = .5. That means n = ((1.96)1. 5 /.5)^2 = 34. 6. Thus n needs to be greater than 36.
(b) Let’s decide that we use n as your answer in part (a). As always, we will use the sample X to estimate the true mean μ. What is the probability that the value of X will be within ±.5 from μ?
Answer: If we use n = 36, then by design of the 95% CI,
P ( X¯ will be within μ ± .5) =. 95.
(c) Suppose n = 20 was used, instead of your answer in part (a). One-sample z-test is about to be performed with H 0 : μ = 120 and HA : μ > 120, and significance level .05. What is the power of the test if μ was actually 121 ounces?
Answer: Since we are looking at upper tailed alternative, P ower = 1 − Φ(zα − √ n(μ − μ 0 )/σ) = 1 − Φ(1. 65 −
(d) In the above one-sample z-test with significance level of .05, if we want to be able to reject the H 0 when μ is actually 121 ounces with probability of .95, what sample size do we need?
Answer: This means that we need 1 − Φ(1. 65 − √n(121 − 120)/ 1 .5) = .95 or Φ(1. 65 − √n(121 − 120)/ 1 .5) = .05. Since the table says Φ(− 1 .65) =. 05 , we know that 1. 65 −
n(121 − 120)/ 1 .5 = − 1 .65, or
n(121 − 120)/ 1 .5 = 3.30. Solve for n, and we get 24.5. So the answer is n=25.
- Wrinkle recovery angle is one of the most important characteristics for evaluating the performance of cross linked cotton fabric. Ester carboxyl band absorbence is known to improve the wrinkle resistance. Below is a data taken for x = absorbance and y = wrinkle resistance angle.
i 1 2 3 4 5 6 7 8 9 10 11 xi .115 .126 .183 .246 .282 .344 .355 .452 .491 .554. yi 334 342 355 363 365 372 381 392 400 412 420
Sample mean and sample standard deviation of x were 0.345 and 0.176, respectively. Sample mean and sample standard deviation of y were 376 and 27.8, respectively. Sample correlation of x and y was 0.992. Below is the output from statistical software.
estimated intercept: 321.9 standard error for the intercept estimator: 2. estimated slope: 156.7 standard error for the slope estimator: 6. estimated variance for error term: 3.
(a) Write the equation for fitted regression line.
Answer: Equation for fitted line is Yˆ = 321.9 + 156. 7 X.
(b) What is the coefficient of determination r^2?
Answer: Since r = .992, r^2 =. 9849.
(c) Test if slope parameter β 1 is zero with significance level .05. What does this mean?
Answer: t = (^1566). 464.^7 = 24.2. P-value ≈ 0. Reject H 0 : β 1 = 0.
(d) Give 95 % CI for mean of wrinkle resistance when x = .5.
Answer:
Yˆ ± tα 2 ,n− 2 σˆ
n +
(x − x¯)^2 SXX^ =^400.^25 ±^ (2.262)3.^605
(. 5 − .345)^2
Note that SXX = (n − 1)S x^2 = .309.
(e) Give 95 % Prediction Interval for next measurement of wrinkle resistance angle when x = .5.
Answer:
Yˆ ± tα 2 ,n− 2 σˆ
1 +^1
n
2 SXX
1 +^1
+ (.^5 −^ .345)
2
. 309 = (391. 4 , 409 .1)
Note that SXX = (n − 1)S x^2.
