Name: ID: Sec: Answer ALL Questions (Full Mark 30 points)

Cairo University

Faculty of Engineering

Systems & Biomedical Eng. Department

Third Year – Term I

SBE 304 - Biostatistics

Mid-Term Exam

Fall 2019

Date: 19 November 2019

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Time Allowed: 90 Minutes

Name: ID: Sec:

Answer ALL Questions (Full Mark 30 points)

Question 1 ( 6 Points ):

1. Suppose the time it takes a data collection operator at a hospital to fill out a patient

electronic record/form for a database is uniformly between 1.5 and 2.2 minutes.

a. What is the mean and variance of the time it takes an operator to fill out the

form?

E(X) = (1.5+2.2)/2 = 1.85 min

V(X) =(2.2 – 1.5)2/12 = 0.0408 min2

b. What is the probability that it will take less than two minutes to fill out the

form?

P(X < 2) = ∫𝟏

(𝟐.𝟐−𝟏.𝟓)

𝟐

𝟏.𝟓 𝒅𝒙 = ∫ 𝟏/𝟎. 𝟕

𝟐

𝟏.𝟓 𝒅𝒙 = 𝟎. 𝟕𝟏𝟒𝟑

c. Determine the cumulative distribution function of the time it takes to fill out the

form.

F(x) = ∫𝟏

(𝟐.𝟐−𝟏.𝟓)

𝒙

𝟏.𝟓 𝒅𝒙 = ∫ 𝟏/𝟎. 𝟕

𝒙

𝟏.𝟓 𝒅𝒙 𝒇𝒐𝒓 𝟏. 𝟓 < 𝒙 < 2.2, 𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆,

2.2,1

,2.25.1,7.0/)5.1(

,5.1,0

)(

x

xx

x

xF

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Mid-Term Exam

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2. The random variable X has the probability distribution given in the table

below.

x 0 1 2 3

P(X = x) 0.2 0.3 u 2v

a. Given that E(X) = 1.6, what is the value of u and v?

0.5 + u + 2v = 1

0.3 + 2u + 6v = 1.6

u = 0.2 v = 0.15

b. Find P( -4 < X ≤ 2)

0.7

c. Find E(5 – 2X)

E(Y) = 5 -2 E(X) = 5 – 2*1.6 = 1.8

d. Find Var(5 – 2X)

Var(X) = (1)2*(0.3) +22 *(0.2) + 32*(0.3) – 1.62= 1.24

Var(Y) = (-2)2 Var(X) = 4.96

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Mid-Term Exam

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3. The measure of intelligence, IQ, of a group of students is assumed to be normally distributed with mean 100 and standard deviation 15.

a. Find the probability that a randomly selected student has an IQ less than 91

P(X < 91) = P(Z < ((91 -100)/15)) = P(Z < -0.6) = 0.2743

b. The probability that a randomly selected student has an IQ of at least 100 + k is 0.2090. Find, to the nearest integer, the value of k

P(X > (100 + k)) = P(Z > ((100 + k - 100)/15)) = P(Z > k/15) = 0.2090

From the tables: k/15 = 0.81 k=12.15 ~= 12

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Mid-Term Exam

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4. A drug company manufactures antibiotics in powder form. Bottles are produced with a mean of 510 mg of powder. Quality control checks show that 1% of bottles are rejected because their weight is less than 485 mg.

a. Find the standard deviation of the weight.

P(X < 485) = P(Z < ((485 -510)/σ)) = 0.01 (485 -510)/σ = -2.3263 σ = 10.746 ~= 10.7

b. Find the proportion of bottles that weigh more than 525 mg.

P(X > 525) = P(Z > ((525 -510)/10.7)) = P(Z > 1.40) = 1- 0.9192 = 0.0808 ~= 8%

c. If the mean weight is changed so that 1% of bottles contain less than the

stated weight of 510 mg. find the new mean weight of the powder.

P(X < 510) = 0.01 (510 – μ)/ 10.7 = -2.32 μ = 534.8 ≈ 535

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Mid-Term Exam

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5. The drug Lipitor is meant to lower cholesterol levels. In a clinical trial of 863 patients who received 10 mg doses of Lipitor daily, 47 reported a headache as a side effect.

a. Obtain a point estimate for the population proportion of Lipitor users who will experience a headache as a side effect. (One Point)

The point estimate for the population proportion = 47/863 = 0.054

b. Verify that the requirements for constructing a confidence interval about this proportion are satisfied. (One Point)

n= 863 p=0.054 np = 46.6 > 5 and n(1-p) = 816.4 > 5

c. Construct a 90% confidence interval for the population proportion of Lipitor users who will report a headache as a side effect. (3 Points)

d. Interpret the confidence interval. (One Point)

We are 90% confident that the proportion of Lipitor users who will experience a headache as a side effect is between 0.042 and 0.067

0.054 – 1.65*sqrt(0.054 (1- 0.054)/ 863) ≤ p ≤ 0.054 + 1.65*sqrt(0.054 (1- 0.054)/ 863)

0.042 ≤ p ≤ 0.067

Cairo University





Mid-Term Exam

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6.

a. Suppose you lust received a shipment of 10 ECG devices. One of them is

defective. You will accept the shipment if two randomly selected devices work.

What is the probability that you will accept the shipment? (One Point)

P = (9/10) * (8/9) = 0.8

OR

It is a hypergeometric random variable where:

N=10 K = 9 n=2

b. A six-character ECG device code is designed so that the first two characters are

English letters and the last four characters are digits (0 through 9). (3 Points)

i. How many different codes can be formed?

ii. If a letter or a digit can be used just once in each code, how many

different codes can be formed?

iii. What is the probability of correctly guessing a certain code?

i- 26 * 26 * 10 * 10 * 10 * 10 = 6760000

ii- 26 * 25 * 10 * 9 * 8 * 7 = 3276000

iii- 1/3276000

Best of Luck!

p(x) =

p(x) = (9*8)/2/(10*9/2) = 36/45 = 0.8

Name: ID: Sec: Answer ALL Questions (Full Mark 30 points)

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