Midterm Scores Min 1Q Median 3Q Max 23 58 68 83 100 The - - PowerPoint PPT Presentation

Midterm Scores

Min 1Q Median 3Q Max 23 58 68 83 100 The exam will be curved, with roughly 70–100 being in the A/B range Scores less than 40 are a matter of concern Students who do substantially better on the final will have their grade based mainly on the final and the homework, otherwise the grade will be based on all three components

October 28, 2016 1 / 21

Paths to Improvement

Come to class Go to section Come to office hours Talk to me Read Lecture 1, slides 16–18, “How to Succeed in this Class” Keep up with the work. You can’t catch up on weeks of class in a couple of days of cramming

October 28, 2016 2 / 21

Problem 1

One step in the manufacture of a medical device involves a fermentation process. A starter solution of engineered yeast cells has an estimated density of λ = 100,000 cells per ml (one milliliter = 1 thousandth of a liter). (a) Suppose that a sample of 1µl (one micro liter = 1 millionth of a liter) is taken of the starter solution and let X be the random variable defined as the number of yeast cells in the sample. Find the mean and the standard deviation of X. Find the coefficient of variation (CV) of X, defined as the ratio of the standard deviation to the mean.

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One µl is 1/1000 of a ml, so the mean number of cells is cut by a factor of 1,000. Also, the distribution will be at least approximately Poisson, implying that the variance is equal to the mean. E(X) = V (X) = 100 SD(X) = 10 CV (X) = 10/100 = 0.1

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(b) Suppose that the process needs that the CV of the sample number X be 0.05. How many microliters should the sample be to achieve this level? Suppose we sample k microliters and Y is the number of yeast cells E(Y ) = V (Y ) = 100k SD(Y ) = 10 √ k CV (Y ) = 10 √ k/100k = 0.1/ √ k = 0.05 √ k = 2 k = 4

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(c) If the sample volume is reduced to 0.1 µl, what is the chance that there are no cells in the sample? The mean is reduces by an additional factor of 10, down to 10. The Poisson pdf is Pr(X = x) = e−1010x/x! so P(X = 0) = e−10100/0! = e−10 = 0.0000454

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Problem 2

Batches of fibroblasts are treated to induce conversion to iPSC’s. A sample of 9 batches has a number of successfully converted cells as follows: {3, 7, 5, 2, 4, 6, 8, 6, 4}. (a) Find the mean, variance, and standard deviation of the number of converted cells.

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• x = 45

¯ x = 45/9 = 5

• (x − ¯

x)2 = 30

• x2 = 255
• (x − ¯

x)2 = 255 − 9(5)2 = 255 − 225 = 30 s2 = 30/8 = 3.75 s = √ 3.75 = 1.936

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(b) Find the five-number summary (using the definitions from the book). What is the IQR? Ordered data: {2, 3, 4, 4, 5, 6, 6, 7, 8}. Ranks of quartiles are 2.5, 5, and 7.5 using f (n + 1). So the five number summary is {2, 3.5, 5, 6.5, 8}. The IQR is 6.5 − 3.5 = 3

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(c) Find the inner fences. Where are the ends of the whiskers in a boxplot? Lower inner fence = 3.5 − (1.5)(3) = −1 Upper inner fence = 6.5 + (1.5)(3) = 11 The ends of the whiskers are at 2 and 8.

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Problem 3

A medical device company uses the following acceptance sampling procedure: Take a sample of size five from the (very large) shipment. Let X be the number of defects in the sample. If there are any defects in the sample (X > 0), then reject the shipment; otherwise accept the shipment. (a) If the true probability is p = 0.1 that a random item in the shipment is defective, find the probability that the shipment is rejected.

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(a) If the true probability is p = 0.1 that a random item in the shipment is defective, find the probability that the shipment is rejected. Pr( Rejected|p = 0.1) = Pr(X > 0) = 1 − Pr(X = 0) = 1 − (0.9)5 = 0.4095 (b) If the true probability is p = 0.2 that a random item in the shipment is defective, find the probability that the shipment is rejected. Pr( Rejected|p = 0.2) = Pr(X > 0) = 1 − Pr(X = 0) = 1 − (0.8)5 = 0.6723

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(c) Suppose that previous experience suggests that there is a 50% chance that a random shipment has p = 0.1 vs. p = 0.2. Find P(p = 0.1|Shipment Rejected). (Hint: first find P(p = 0.1 ∩ Shipment Rejected) and P(p = 0.2 ∩ Shipment Rejected), then P(Shipment Rejected), and finally P(p = 0.1|Shipment Rejected) using the laws of probability.)

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Pr(p = 0.1 ∩ Rejected) = Pr( Rejected|p = 0.1)Pr(p = 0.1) = (0.4095)(0.5) = 0.2048 Pr(p = 0.2 ∩ Rejected) = Pr( Rejected|p = 0.2)Pr(p = 0.2) = (0.6723)(0.5) = 0.3362 Pr( Rejected) = 0.2048 + 0.3362 = 0.5409 Pr(p = 0.1| Rejected) = Pr(p = 0.1 ∩ Rejected)/Pr( Rejected) = 0.2048/0.5409 = 0.3786

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Problem 4

A piece of raw material for a medical device has an 80% chance of having no defects, a 15% chance of having 1 defect, and a 5% chance of having 2 defects. Let X be the random variable of the number of defects. (a) Find the mean, variance, and standard deviation of X.

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(a) Find the mean, variance, and standard deviation of X. x P(x) xP(x) x2P(x) 0.80 1 0.15 0.15 0.15 2 0.05 0.10 0.20 1.00 0.25 0.35 µX = 0.25 σ2

X

= 0.35 − (0.25)2 = 0.2875 σX = 0.5362

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(b) If each defect costs $5 to fix, let Y be the random variable of the cost of repair of the piece of raw

• material. Find the mean, variance, and standard

deviation of Y . µY = (5)(0.25) = $1.25 σ2

Y

= (5)2(0.2875) = $7.19 σY = (5)(0.5362) = $2.68

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(c) Suppose Xi (i = 1, 2, . . . , 20) are the numbers of defects in 20 pieces of raw material, assumed to be statistically independent. Find the mean, variance, and standard deviation of the total number of defects W in the 20 pieces. µW = (20)(0.25) = 5 σ2

W

= (20)(0.2875) = 5.75 σW = √ 5.75 = 2.398

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Problem 5

Suppose that X is a uniform random variable, meaning that it has the pdf f (x) =

• 1

0 ≤ x ≤ 1

• therwise

(a) Find the mean, variance, and standard deviation of X.

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(a) Find the mean, variance, and standard deviation of X. µX = 1 xf (x) dx = 1 x dx = 1

2x2

• 1

= 0.5 1 x2 dx =

1 3x3

• 1

= 1

3 = 0.3333

σ2

X

=

1 3 − 0.52 = 1 12 = 0.0833

σX = √ 0.0833 = 0.289

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(b) Find the mean, variance, and standard deviation of the average Y of X1, X2, . . . X10, where each of Xi has the pdf as above, and they are statistically independent. µY = µX = 0.5 σ2

Y

= σ2

X/10 = 0.00833

σY = σX/ √ 10 = 0.0913

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