Worksheet week 49

AP Stats course
Lecturer: Hans van der Zwan

Textbook
Starnes D. S., et al. (2015). The Practice of Statistics (5th ed.). New York: W. H. Freeman and Company/BFW.

Worksheet Thu 2021-12-09

First name: …………………………………….
Last name: ……………………………………..

Exercises on this worksheet should be worked out thoroughly.
Indicate clearly the methods you use, because on exams you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations.
You can use the examples to write it up correctly.

Example 1

Bottles of milk should contain 1,000 ml.
If the filling process is under control, the average content of the bottles is 1,380 ml with a standard deviation of 10 ml. The contents of the bottles are approximately normally distributed.

Assume the filling process is under control.

What is the probability that a random chosen bottle contains more than 1,015 ml? Answer in three decimals.
Someone buys 12 bottles; assume the contents of the different bottles are independent?
What is the probability that the total content of the 12 bottles is more than 12,120 ml. Answer in three decimals.

Someone buys two bottles.

What is the probability that the total content of these two bottles is more than 2,030 ml? Answer in three decimals.
What is the probability that both bottles contain more than 1,015 ml? Answer in three decimals.

Answer
X: the content of a randomly chosen bottle X ~ N(\(\mu\) = 1,000, \(\sigma\) = 10) ml

P(X > 1,015) = 0.067 (with TI 84)

T: total contents of 12 bottles

T = X₁ + X₂ + … + X₁₂; with X_i: contents of the \(i^{th}\) bottle.
\(\mu_T\) = 1,000 + 1,000 + … + 1,000 = 12 \(\times\) 1,000 = 12,000
\(\sigma_T\) = \(\sqrt{10^2 + 10^2 + ... + 10^2}\) = \(\sqrt{12 \times 10^2}\) = 10\(\sqrt{12}\)
T ~ N(\(\mu\) = 10,150, \(\sigma\) = 10\(\sqrt{12}\))
P(T > 12,120) = 0.00027 \(\approx\) 0.000 (TI 84); note the corresponding z-value is 3.46.

D: total content of two bottles
D ~ N(\(\mu=2,000\); \(\sigma = 10\sqrt{2}\))

P(D > 2030) = 0.017
P(both bottles contain more than 1,015 ml) = P(X₁ 1,015 AND X₂ 1,015) = [X1 and X2 are independent] = P(X₁ > 1,015) \(\times\) P(X₂ > 1,015) = 0.067 \(\times\) 0.067 \(\approx\) 0.004

Exercise W1.1
According to the producer of a certain brand of light bulbs the burning time is approximately normally distributed with an average of 12,000 hours and a standard deviation of 150 hours.
Assume the information from the producer is valid.

What is the probability that a randomly chosen light bulb has a burning time of less than 12,000 hours?
\[\\[1.5in]\]
What is the probability that a randomly chosen light bulb has a burning time of less than 11,900 hours? \[\\[1.5in]\]

An inspector of a consumer organization buys 5 light bulbs to test them.

What is the probability that the average burning time of the five light bulbs is less than 12,000 hours?
\[\\[1.75in]\]
What is the probability that the average burning time of the five light bulbs is less than 11,900 hours? \[\\[1.5in]\]
What is the probability that all five of them have a burning time of less than 12,000 hours? \[\\[1.5in]\]
If the average burning time of the five light bulbs is less than 11,900 hours, would you still believe the producer’s claim? Explain. \[\\[1in]\]
If all five tubes have a burning time of less than 12,000 hours, would you still believe the claim of the producer? Explain.
\[\\[1.5in]\]

Example 2

From a large batch of ballpoint pens 4% is defective.
A random sample of 10 pens is drawn from this batch.

What is the probability that none of these pens are defective?
What is the probability two of these pens are defective?
What is the probability that two or less of these 10 pens are defective?
What is the probability that two or more of these 10 pens are defective?
If a random sample of 50 pens is drawn, what is the expected value of the number of defective pens in such a sample?
What is the meaning of the expected value calculated in part (v)?

Answer
K: number of defective pens in the sample K ~ bin(n=5; p=0.04)

P(K=0) = \(0.96^5\) = 0.8154
or

P(K=0) = \(\binom{4}{0}\times 0.04^0 \times 0.96^5 = \frac{4!}{0!\times(4-0)!}\times 0.04^0 \times 0.96^5 = 0.8154\)

or use the TI84

P(K=2) = \(\binom{4}{2}\times 0.04^2 \times 0.96^3 = \frac{4!}{2!\times(4-2)!}\times 0.04^2 \times 0.96^3 = 0.0142\)

or use TI84

P(K <= 2) = P(K = 0) + P(K = 1) + P(K = 2) = ……
smarter to use TTI84 P(K <= 2) = 0.9993
P(K >= 2) = 1 - P(K < 2) = 1 - P(K <= 1) = 1 - … = …
L: number of defective pens in a random sample of 50 pens
L ~ bin(n = 50, p = 0.04)
E(L) = np = 50 x 0.04 = 2
If many random samples of 50 pens are drawn, the average number of defective pens in these random samples will be close to 2 (the more samples, the closer, that is because of the law of large numbers)

Exercise W1.2
A teacher at u university teaches a course to more than 800 hundreds of students. He states that more than 80% of his students is satisfied about his teaching. Assume that the percentage of satisfied students equals 80%.
To measure the student satisfaction, a random sample of 40 students is drawn and asked to fill in a questionnaire about the course.

What is the expected value of the number of students that are satisfied about the teaching of this course? \[\\[1.5in]\]
What is the probability that 32 students in the sample are satisfied about the teaching? \[\\[1.5in]\]
What is the probability that less than 25 students in the sample are satisfied about the teaching of this teacher? \[\\[1.5in]\]
What is the probability that more than 37 students in the sample are satisfied about the teaching of this teacher? \[\\[1.5in]\]

Suppose that no random sample is taken, but that all students are asked to complete the questionnaire and only 40 have completed the questionnaire. Only 20 of them were satisfied about the teaching. The management confronted the teacher with this result and told him that only 50% of the students is satisfied about his teaching.

What argument can the teacher use in his favor? \[\\[1.5in]\]