Practice Problems for Homework #6. Normal distribution and Central Limit Theorem.
1. Read Section 3.4.6 about the Normal distribution and Section 4.7 about the Central Limit Theo-rem.
2. Solve the practice problems below.
3. Open Homework Assignment #6, solve the problems, and submit multiple-choice answers. 1. (10 marks) Let Z be a Standard Normal random variable.
Compute (a) P (Z < 0.97) (b) P (Z ≤ 0.97) (c) P (Z > 0.97) (d) P (|Z| ≤ 0.97) (e) P (Z < −5.0) (f) P (Z > −5.0)
2. (10 marks) Let X be a Normal random variable with parameters µ = 8 and σ = 3.5. Compute
(a) P (X < 0) (b) P (X > 15) (c) P (|X| < 2)
3. (10 marks) (Sec 3.4 (page 147), #1.)
Lifetimes of VLSI chips manufactured by a semiconductor manufacturer are approximately Nor-mally distributed with µ = 5 × 106 h and σ = 5 × 105 h. A computer manufacturer requires that at least 95% of a batch should have a lifetime greater than 4 × 106h. Will the deal be made? 4. (10 marks) TravelByUs is an internet-based travel agency wherein the customers can see videos
of the cities they plan to visit. The number of daily hits at its website is Normally distributed with µ = 1, 000 and σ = 240.
(a) What is the probability of getting fewer than 900 hits?
(b) The website of this company has a limited bandwidth, which is measured in terms of the number of hits the site can handle. How large a bandwidth should TravelByUs have in order to handle 99% of the daily traffic?
5. (10 marks) The scoring of modern IQ tests is such that Intelligence Quotients (IQs) have Normal distribution with µ = 100 and σ = 15.
(a) What percent of people have IQ less than 80? (b) What percent of people have IQ greater than 120?
(c) Mensa International is a non-profit organization that accepts only people with IQ within the top 2%. What level of IQ qualifies one to be a member of Mensa?
6. (10 marks) The time spent by students working on a project is a Normal random variable with parameters µ = 12 hours and σ = 4 hours. Compute the probability that
(a) the amount of time spent (b)on the project is less than 14 hours. (b) amount of time spent on the project is greater than 8 hours.
7. (10 marks) A keyword search program lists the files that contain a given keyword. If it runs through 200 files, and each file contains the keyword with probability 0.36, independently of other files, compute the probability that
(a) more than 70 files (b) less than 70 files
(c) exactly 70 files will be listed.
8. (10 marks) Among all the computer chips produced by a certain factory, 6 percent are defective. A sample of 400 chips is selected for inspection.
a) What is the probability that this sample contains between 20 and 25 defective chips (includ-ing 20 and 25)?
b) Suppose that each of 40 inspectors collects a sample of 400 chips. What is the probability that at least 8 inspectors will find between 20 and 25 defective chips in their samples? 9. (10 marks) The homework consists of 40 independent problems. On the average, it takes 5
minutes to solve a problem, with a standard deviation of 2 minutes. Find the probability that the homework will be completed in less than 3 hours.
10. (10 marks) Seventy independent messages are sent from an electronic transmission center. Mes-sages are processed sequentially, one after another. Transmission time of each message is Expo-nential and with a mean of 0.2 minutes. Find the probability that all 70 messages are transmitted in less than 12 minutes.
11. (10 marks) An average scanned image occupies 0.6 megabytes of memory with a standard devi-ation of 0.4 megabytes. If you plan to install 80 images on your web site, what is the probability that their total size is between 47 megabytes and 56 megabytes.
12. (10 marks) The installation time of a certain software follows Gamma distribution with param-eters r = 100 and λ = 2 min−1.
a) Compute the probability that it takes at least 1 hour to install this software.
Solutions:
1.
From the Table of Normal distribution, (a) P (Z < 0.97) = F (0.97) = 0.8340 (b) P (Z ≤ 0.97) = F (0.97) = 0.8340
(c) P (Z > 0.97) = 1 − P (Z < 0.97) = 1 − 0.8340 = 0.1660
(d) P (|Z| ≤ 0.97) = F (0.97) − F (−0.97) = 0.8340 − 0.1660 = 0.6680
(e) P (Z < −5.0) = 0.0000 (very small area under the density curve to the left of 5.0) (f) P (Z > −5.0) = 1.0000 (virtually the entire area under the density curve)
2.
Standardize X and use the Table of Normal distribution. (a) P (X < 0) = P (Z < 0−83.5) = P (Z < −2.29) = 0.011 (b) P (X > 15) = P (Z > 15−83.5 ) = P (Z > 2) = 0.023
(c) P (|X| < 2) = P (−2 < X < 2) = P (−2.86 < Z < −1.71) = 0.044 − 0.002 = 0.042 3.
Here X has a normal distribution with µ = 5 × 106h and σ = 5 × 105h.
P (X > 4 × 106) = P (Z > −2) = 0.977 (using the Normal distribution table and symmetry).
The deal will be made since this probability is more than 0.95. 4.
Let X be the number of hits at the website, which is is Normal (µ = 1000, σ = 240). With a continuity correction (because the number of hits is integer),
(a) P (X < 900) = P (X < 899.5) = P (Z < (899.5 − 1000)/240) = P (Z < −0.42) = 0.34. (b) We need k such that
0.99 = P (X ≤ k) = P (X ≤ k + 0.5) = P (Z ≤ (k + 0.5 − 1000)/240).
From the Normal table, (k − 999.5)/240 = 2.33, which means k = 2.33(240) + 999.5 = 1559. 5.
(b) P (X > 120) = P (Z > 120−10015 ) = P (Z > 1.33) = 1 − 0.9082 = 0.0918 or 9.18% [also follows by symmetry between (a) and (b)]
(c) Solve the equation P (X ≥ x) = 0.02 for x.
Look for the probability 1 − 0.02 = 0.98 in the Table of Normal distribution and find the corre-sponding value of z = 2.05. Then, from
P (X ≥ x) = P Z ≥ x − 100 15 = 0.02, we have x−10015 = 2.05, and x = (2.05)(15) + 100 = 130.75.
Members of Mensa should have IQ starting at 130.75. 6.
The time X has Normal distribution with µ = 12 and σ = 4. Let Z = (X − µ)/σ denote a Normal(0, 1) random variable.
From the Normal table,
(a) P (X < 14) = P (Z < (14 − 12)/4) = P (Z < 0.5) = 0.69
(b) P (X > 8) = P (Z > (8 − 12)/4) = P (Z > 1) = 1 − 0.84 = 0.16 7.
Let X be the number of files listed. Then X is Binomial(n = 200, p = 0.36). Since n is large and p is neither too small nor too large, the distribution of X is approximately Normal with parameters
µ = 200 · 0.36 = 72 and σ2 = 200 · 36 · (1 − 0.36) = 6.792. Applying the continuity correction and using the table of Normal distribution, (a) P (X > 70) = P (X > 70.5) = P (Z > −0.22) = 0.5871
(b) P (X < 70) = P (X < 69.5) = P (Z < −0.37) = 0.3557
(c) P (X = 70) = P (X < 70.5) − P (X < 69.5) = (Z < −0.22) − P (Z < −0.37) = 0.0572. 8.
a) Let X be the number of defective chips in the sample of 400. Then, X ∼ Binomial (n = 400, p = 0.06). Approximately, X ∼ Normal (µ = 400(0.06) = 24, σ2 = 400(0.06)(1 − 0.06) = 4.752). Applying the continuity correction and using the normal table,
b) Let Y be the number of inspectors who find between 20 and 25 defective chips in a sample of 400. Using (a), Y ∼ Binomial (n = 40, p = 0.45). Approximately, Y ∼ Normal (µ = 40(0.45) = 18, σ2 = 40(0.45)(1−0.45) = 3.152). Applying the continuity correction
and using the normal table
P (Y ≥ 8) = P (Y > 7.5) ≈ P (Z > −3.33) = 1.
9.
Let T denote the total time (in minutes) it takes to do the homework. From the central limit theorem, T approximately follows a Normal (E(T ) = 40(5) = 200, var(T ) = 40(22) = 12.652)
distribution. Hence from the normal table,
P (T < 180) ≈ P (Z < (180 − 200)/12.65) = P (Z < −1.58) = 0.06. 10.
Let Xi denote the transmission time (in minutes) of the ith message, i = 1, . . . , 70. It is given
that the Xi follow independent exponential distributions with E(Xi) = 0.2. It follows that λ =
1/0.2 = 5 and hence var(Xi) = 1/52 = 0.22. Let T = X1+ . . . + X70be the total transmission
time. From the Central Limit Theorem, T approximately follows a Normal distribution with E(T ) = 70(0.2) = 14 and var(T ) = 70(0.22) = 1.672. Hence
P (T < 12) ≈ P (Z < (12 − 14)/1.67) = P (Z < −1.20) = 0.115, using the normal table.
11.
Let Xi denote the size (in megabytes) of the ith scanned image, i = 1, . . . , 80. It is given
that E(Xi) = 0.6 and var(Xi) = 0.42. Let T = X1 + . . . + X80 be the total size of 80
images. From the Central Limit Theorem, T approximately follows a Normal distribution with E(T ) = 80(0.6) = 48 and var(T ) = 80(0.42) = 3.582. Hence, using the normal table,
P (47 < T < 56) ≈ P (Z < 2.23) − P (Z < −0.28) = 0.99 − 0.39 = 0.60, 12.
The installation time X has Gamma distribution with µ = r/λ = 50 min and σ = pr/λ2 = 5
b) P (30 ≤ X ≤ 60) = P 30−505 ≤ Z ≤ 60−50
5 = P (−4 ≤ Z ≤ 2) = F (2) − F (−4) =
