Exam 2 - Math 3200 Spring 2013

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General instructions: This exam has 13 questions, each worth the same amount. Check that no pages are missing and promptly notify one of the proctors if you notice any problems with your copy of the exam. Mark your ID number on the six blank lines on the top of your answer card using one line for each digit. Print your name on the top of the card. Choose the answer closest to the solution and mark your answer card with a PENCIL by shading in the correct box. You may use a 4× 6 card with notes and any calculator that does not have graphing functions. GOOD LUCK!

1. An office supply warehouse receives an order for three computers. The warehouse has 50 computers in stock, two of which are defective. The order is filled by randomly drawing from the computers in stock. Let X be the number of defective computers in the order. What probability distribution describes P (X = x)? (A) Gamma (B) Uniform (C) Geometric (D) Bernoulli (E) Binomial (F) Exponential (G) Poisson (H) Hypergeometric (I) Beta (J) Normal Solution: This is a hypergeometric distribution:

P (X = x) =

¡ 2¢¡ x

48 ¢ 3−x ¡50¢ . 3

2. (Continuation of the previous problem.) Assume the same information given in problem 1. What is the probability P (X = 1)? (A) 0.920 (B) 0.713 (C) 0.545 (D) 0.432 (E) 0.316 (F) 0.115 (G) 0.089 (H) 0.055 (I) 0.032 (J) 0.012 Solution:

¡2¢¡48¢

P (X = 1) = 1¡502¢ =

2 × 48×47 2

3

2

50×49×48 3×2

=

141 = 0.1151. 1225

3. A car travels between two cities A and C which are 75 miles apart. If the car has a breakdown, the distance X from the breakdown to city A is distributed as U [0, 75]. The driver is a member of an automobile service that has contracts with garages in cities A, B , and C , where city B is between cities A and C , 25 miles from city A. If the car breaks down, it is towed to the closest garage. Find the probability that the car is towed more than 10 miles given that it is more than 20 miles from city A. (A) 0.725 (B) 0.255 (C) 0.450 (D) 0.300 (E) 0.654 (F) 0.545 (G) 0.575 (H) 0.467 (I) 0.324 (J) 0.628 Solution: It is easily seen that the function D(x) representing the distance that the car needs to be towed is:

   x      25 − x D(x) =   x − 25      75 − x

if x ≤ 25/2 if 25/2 ≤ x ≤ 25 if 25 ≤ x ≤ 50 if 50 ≤ x ≤ 75

An inspection of the graph of D(x) (see below) shows that P (D ≥ 10 |breakdown more than 20 miles from A) =

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length([35, 65]) 30 = = 0.545. 75 − 20 55

4. Let X be the time to failure of a light bulb. Assume that X is exponentially distributed. If the mean time to failure is 10, 000 hours, what is the median time to failure? (A) 6531.5 (B) 5735.5 (C) 4339.2 (D) 6297.3 (E) 5942.6 (F) 7631.1 (G) 2730.8 (H) 3358.4 (I) 9351.5 (J) 6931.5 Solution: We know that the mean value of X ∼ Exp(λ) is 1/λ, so the rate parameter is λ = 1/10000.

We also know that the cumulative distribution function is F (x) = 1 − e −λx , and that the median (0.5quantile) is the solution to 0.5 = F (x). It follows that

median = −10000 log 0.5 = 6931.5

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5. (Continuation of the previous problem.) Assume the same information given in problem 4. What is the probability that the bulb will last an additional 1000 hours if it has already lasted for 1000 hours? (A) 0.675 (B) 0.320 (C) 0.459 (D) 0.197 (E) 0.883 (F) 0.954 (G) 0.905 (H) 0.225 (I) 0.991 (J) 0.392 Solution: By the memoryless property of exponential random variables, P (X ≥ 2000 |X ≥ 1000 ) = P (X ≥ 1000) = 1 − F (1000) = e −1000/10000 = e −1/10 = 0.905.

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6. The weight of coffee in a can is normally distributed with a mean of 16.1 oz. and standard deviation of 0.5 oz. What is the probability that a can contains 16 to 16.5 oz. of coffee? (A) 0.617 (B) 0.682 (C) 0.574 (D) 0.521 (E) 0.486 (F) 0.449 (G) 0.398 (H) 0.367 (I) 0.225 (J) 0.292 Solution: If W ∼ N (16.1, 0.52 ), then P (16 ≤ W ≤ 16.5) = P

µ

¶ 0.4 −0.1 = P (Z ≤ 0.8) − P (Z ≤ −0.2) = P (Z ≤ 0.8) − [1 − P (Z ≤ 0.2)] . ≤Z ≤ 0.5 0.5

Looking up at the table of the standard normal c.d.f., P (Z ≤ 0.2) = 0.579 and P (Z ≤ 0.8) = 0.788. There-

fore,

P (16 ≤ W ≤ 16.5) = 0.788 + 0.579 − 1 = 0.367.

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0.4 0.0

0.2

density

0.6

0.8

7. A large sample is taken from the asymmetric distribution whose density is shown below.

−4

−3

−2

−1

0

x

Decide whether each of the following statements is true (T) or false (F). (a) The mean is likely to be larger than the median; (b) The sample is likely to be negatively skewed; (c) The third quartile Q 3 is likely to be farther from the median than the first quartile Q 1 . (A) F T F (B) T T F (C) F T T (D) T F F (E) T T T (F) F F T (G) T F T (H) F F F Solution: The distribution is clearly negatively skewed, so the mean should be smaller than the median and the first quartile should the farther from the median than the third quartile. A large sample is likely to share all these properties.

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8. Ten observations of a certain random variable produced the following set of numbers: 5.3 6.3 6.4 6.5 6.7 6.9 7.0 7.6 7.9 8.8. Determine the interquartile range for this data. (A) 2.0 (B) 1.9 (C) 1.8 (D) 1.7 (E) 1.6 (F) 1.5 (G) 1.4 (H) 1.3 (I) 1.2 (J) 1.1 Solution: Recall that the 100pth percentile is determined by x˜p = x (m) + (p(n + 1) − m)(x (m+1) − x m ) where m is the integer part of (n + 1)p. For the first quartile, (n + 1)p = 11 × 0.25 = 2.75 so m = 2 and x˜ p = x (2) + (2.75 − 2)(x (2) − x 2 ) = 6.3 + 0.75 × (6.4 − 6.3) = 6.375. For the thirdt quartile, (n + 1)p = 11 × 0.75 = 8.25 so m = 8 and x˜ p = x (8) + (8.25 − 8)(x (9) − x 9 ) = 7.6 + 0.25 × (7.9 − 7.6) = 7.675. Therefore the interquartile range is IQR = 7.675 − 6.375 = 1.3.

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9. The correlation coefficient between the midterm scores and final scores in a course is 0.75. A student scored one standard deviation above the mean on the midterm. If the final mean and SD are 80 (out of 100) and 12, respectively, what is that student’s predicted score on the final? (A) 69 (B) 70 (C) 75 (D) 79 (E) 80 (F) 83 (G) 85 (H) 89 (I) 90 (J) 94 Solution: The regression line predicting the final score F in terms of the midterm score M is M −M F −F =r sF sM where (M − M)/s M = 1, F = 80 and s F = 12, and r = 0.75. Therefore, F = 80 + 12 × 0.75 × 1 = 89.

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10. A soft drink company uses a filling machine to fill cans. Each 12 oz. can is to contain 355 milliliters of beverage. In fact, the amount varies according to a normal distribution with mean µ = 355.2 ml and standard deviation σ = 0.5 ml. What is the probability that an individual can contains less than 355 ml?

(A) 0.254 (B) 0.534 (C) 0.432 (D) 0.345 (E) 0.556 (F) 0.412 (G) 0.315 (H) 0.637 (I) 0.763 (J) 0.298 Solution: We want to find P (X < 355) if X ∼ N (355.2, 0.52 ). The z-score for X = 355 is Z=

355 − 355.2 = −0.4. 0.5

Therefore, P (X < 355) = P (Z < −0.4) = 1 − P (z < 0.4) = 1 − 0.6554 = 0.3446.

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11. (Continuation of the previous problem.) Assume the same information given in problem 10. What is the probability that the mean content of a six-pack of cans is less than 355 ml? (A) 0.1635 (B) 0.2153 (C) 0.3205 (D) 0.4015 (E) 0.4883 (F) 0.5567 (G) 0.6078 (H) 0.7832 (I) 0.8550 (J) 0.9325 We are now interested in X = (X 1 +· · ·+X 6 )/6, where the X i ∼ N (355.2, 0.52 ) are i .i .d . random variables.

The mean of X is the same as that of X , which is 355.2, and the standard deviation is p p σ/ n = 0.5/ 6 = 0.2041. Thus P (X < 355) = P (Z
2σ2 ). (A) 0.150 (B) 0.995 (C) 0.975 (D) 0.100 (E) 0.025 (F) 0.005 (G) 0.050 (H) 0.990 (I) 0.500 (J) 0.950 Solution: Note that

and recall that

(n−1)S 2 σ2

¶ µ ¡ ¢ (n − 1)S 2 > 2(n − 1) , P S 2 > 2σ2 = P σ2

∼ χ2n−1 . Therefore, ¢ ¢ ¡ ¡ P S 2 > 2σ2 ≈ P χ2n−1 > 2(n − 1) .

When n = 8,

¡ ¢ ¡ ¢ P S 2 > 2σ2 ≈ P χ27 > 14 = α

where, by definition, χ27,α = 14. Therefore, from the Chi-squared distribution table: α ≈ 0.05.

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13. Consider the student’s t -distribution with 10 degrees of freedom. Find a constant a so that P (|T10 | < a) = 0.95. (A) 3.259 (B) 4.321 (C) 2.228 (D) 6.314 (E) 9.925 (F) 14.089 (G) 1.886 (H) 2.998 (I) 0.261 (J) 0.690 Solution: Observe that P (|T10 | < a) = 1 − 2P (T10 > a) . From this we obtain, P (T10 > a) =

1 − 0.95 = 0.025. 2

Therefore, a = t 10,0.025 = 2.228.

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