Module 24 - Statistics 1 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 24

STATISTICS I

Module Topics 1. Probability 2. Conditional probability 3. Combinations and permutations 4. Discrete and continuous random variables

A:

Work Scheme based on JAMES (THIRD EDITION)

1. All of you should have met some of the work in this module, which is on probability. Turn first to p.858 and read section 13.1, the introduction. 2. Read section 13.2.1 which contains general comments on sampling. Read sections 13.2.2 and 13.2.3 on plotting data, although the topics are not considered further in these modules. 3. Study section 13.3.1 on interpreting probabilities, and then study the first two paragraphs in section 13.3.2. Work through Examples 13.1 and 13.2. The word experiment is used for any process in which the result of each performance depends on chance and therefore cannot be predicted uniquely – e.g. throwing a die and observing the number that appears uppermost, the random selection and testing of 10 fuses from a box containing 100. The result of a single performance of a statistical experiment is called an outcome. Study the remainder of section 13.3.2 on p.865. The notation and interpretations for union, intersection and complement are very important. 4. Before it is possible to discuss the probability of some event, it is necessary to assign a probability to each of the outcomes in the sample space. The probability that is assigned to a given outcome is equal to the proportion of times that the outcome is expected to occur in a large number of repetitions of the experiment. This is known as the frequency interpretation of probability. Suppose an experiment is carried out N times under the same experimental conditions, and that the outcome e1 occurs n1 times. Then n1 /N is called the relative frequency of e1 . The probability, p1 , of e1 occurring is the proportion of times that e1 occurs in a long series of repeated experiments, i.e. p1 = lim (n1 /N ) . N →∞

Study section 13.3.3 on the axioms of probability. All seven numbered results are important and you need to remember them. Note that rule (4) states that p(A) = 1 − P (A) , i.e. the probability an event does NOT occur is 1 minus the probability that it does occur. 5. As mentioned at the bottom of p.866 Venn diagrams are useful in illustrating many of the results. The sample space is represented by a rectangle. The area inside a closed curve within the rectangle represents –1–

an event A, so figure 1a below is a Venn diagram of an event A, showing A and its complement A.

S

11 00 00 11 00 11 A 00 11 00 11

S A

A

A

B

B

figure 1c

figure 1b

figure 1a

S

S

A

B

figure 1d

The shaded area in figure 1b above represents A ∪ B, the union of A and B, i.e. all outcomes that belong to A or B. Note that an outcome in A ∪ B can lie in both A and B. In an analogous way figure 1c above illustrates A ∩ B, the intersection of A and B, with the shaded area this time showing all outcomes in both A and B. When no outcomes lie in both A and B, i.e. A ∩ B = ∅ , (see figure 1d above), then J. calls the events disjoint. It is more usual to say that in this situation the events are mutually exclusive. The general addition rule (7) on p.866 is illustrated on that page by figure 13.7, but is more clearly shown in figure 2 below.

= P(A or B)

+ P(B)

P(A)

P(A and B)

figure 2 Work through Examples 13.3 and 13.4. Note that in Example 13.3 P ({2, 4, 6}) = P (2 or 4 or 6) = P (2) + P (4) + P (6) =

1 1 1 1 + + = , 6 6 6 2

since the events of obtaining a 2, 4 or 6 in a single throw are mutually exclusive. In a similar way it can be shown that P ({1, 3, 5}) = 1/2. In Example 13.4, 80% of students passed the examination in mathematics, and this implies that the probability of any one student passing this examination is 0.8. De Morgan’s law is not necessary for part (c). Instead, you can use the Venn diagram shown as figure 1b above, which allows you to see that the probability that a student fails both mathematics and laboratory work is 1 minus the probability that he passes at least one of them. 6. Study section 13.3.4 on conditional probability on p.868, and then work through Examples 13.5 and 13.6. In part (b) of the latter example, the result is obtained using P (A ∩ D) = 1 − P (A ∪ D) = 1 − {P (A) + P (D) − P (A ∩ D)}, the result mentioned at the end of section 5 above. ***Do Exercises 7, 8, 11 on pp.874 and 875*** 7. It is shown in the upper half of p.870 that the defined conditional probabilities satisfy the axioms of probability, but you do not need to study that section. Look at Example 13.7. The solution given in J. seems rather complicated, and you may find the one below easier to follow. You are given the information that

–2–

P (rain|given previous day sunny) = P (R|S) = 0.6,

P (rain|given previous day rainy) = P (R|R) = 0.8 .

Since every day is either sunny or rainy, it follows that P (S|S) = 1 − 0.6 = 0.4, P (S|R) = 1 − 0.8 = 0.2 . You need to calculate the probability that Saturday is sunny knowing that Thursday was rainy. There are only two possibilities for Friday, that it was sunny or rainy. Hence P (Sat. sunny) = P ((Sat. sunny and Fri. sunny) or (Sat. sunny and Fri. rainy)) = P (Sat. sunny and Fri. sunny) + P (Sat. sunny and Fri. rainy) = P (Sat. sunny|Fri. sunny)P (Fri. sunny) + P (Sat. sunny|Fri. rainy)P (Fri. rainy) = P (Sat. sunny|Fri. sunny)P (Fri. sunny|Thurs. rainy) + P (Sat. sunny|Fri. rainy)P (Fri. rainy|Thurs. rainy) = P (S|S)P (S|R) + P (S|R)P (R|R) = (0.4)(0.2) + (0.2)(0.8) = 0.24. 8. Study section 13.3.5 on independence of events. To help you understand the first paragraph note that, from its definition, P (B1 ) P (B1 and A) , = , P (B1 |A) = P (A) P (A) since if B1 occurs then A always does also (see figure 13.11). The figure also shows that P (A) < 1, therefore P (B1 )/P (A) > P (B1 ). However, B2 and A are mutually exclusive, so P (B2 and A) = 0, and hence P (B2 |A) =

P (B2 and A) = 0, P (A)

< P (B2 ).

J. states that the events A and B are independent if P (B|A) = P (B). It is more usual to express the result in the form A and B are independent if P (A and B) = P (B)P (A). Work through Examples 13.8, 13.9, 13.10 and 13.11. 9. The conditional probability is a particularly powerful tool in the manipulation of probabilities, as this section will demonstrate. Consider this example: Example A: At a plant manufacturing floppy disks for microcomputers, three machines are currently in use. Machine A manufactures 60% of the disks, machine B 30% and machine C 10%. However, only 1% of disks produced on machine A are faulty, whilst 2% are faulty from machine B and 5% from the oldest machine, C. What is the probability that the next disk manufactured at the plant will be faulty? To answer this question, let F be the event that a disk is faulty, and then look at figure 3 below. Note first that S A F B C

figure 3 the sample space S can be subdivided into the three mutually exclusive (and exhaustive) events A, B and C (representing the three machines), since these are the only machines involved in the manufacture and a disk can be made on only one of them. The event F intersects with all three of A, B and C. Consequently, we can write F = (F ∩ A) ∪ (F ∩ B) ∪ (F ∩ C) , –3–

and, since A, B and C are mutually exclusive, so are F ∩ A, F ∩ B and F ∩ C. Thus, taking the probability of this relation gives P (F ) = P (F and A) + P (F and B) + P (F and C) . However, we know P (F and A) = P (A)P (F |A) etc. and, therefore, obtain the result P (F ) = P (A)P (F |A) + P (B)P (F |B) + P (C)P (F |C) . For our Example, all the probabilities on the right-hand side of the equation are known and can be summarised as P (A) = 0.6 P (F |A) = 0.01 P (B) = 0.3 P (F |B) = 0.02 P (C) = 0.1 P (F |C) = 0.05 Consequently, P (F ) = (0.6)(0.01) + (0.3)(0.02) + (0.1)(0.05) = 0.017 . Example A, cont’d: Suppose now that a floppy disk is picked up at random and found to be faulty. What is the probability that it was manufactured on machine A? Answer this by noting P (F and A) = P (A)P (F |A) = P (F )P (A|F ) . Manipulating this expression leads to P (A|F ) =

(0.6)(0.01) P (A)P (F |A) = = 0.353 . P (F ) (0.017)

***Do Exercises 18, 19, 22 on p.875*** 10. For many problems involving probability, the counting of outcomes corresponding to various events becomes difficult unless compact counting methods are developed. A permutation is an ordered arrangement of a set of objects, and the number of permutations of n distinct objects is n! (pronounced ‘ n factorial’), where n! = n(n − 1)(n − 2) · · · 3.2.1. On the other hand, a combination is a subset selected from a set of distinct objects
(without regard to order). The numberofcombinations of size r that can be drawn from n objects is nr (pronounced ‘ n n! n . This is often said to be the number of ways of choosing r from n. choose r ’), where = r!(n − r)! r Example B: Find the number of permutations of the letters a, b and c. Using the formula the answer is 3! = 3.2.1 = 6 , namely abc, acb, bac, bca, cab and cba. Example C: Find the number of ways of choosing two letters from four.   4.3 4! 4 = = 6. Here the formula gives = 2!2! 2.1 2
You should note that in evaluating nr , the calculation is simplified if, in the first instance, the largest factorial in the denominator is divided into the numerator. Thus, for example,   8.7.6 8! 8 = = 56 . = 3!5! 3! 3

–4–

Example D: A box contains 12 valves, four of which are faulty. (i) What is the probability that, if two valves are drawn at random, both are faulty? (ii) What is the probability that at least one of the pair is faulty? (i) The solution is derived here using combinations,
but other methods are possible. The number of ways = 66. These pairs make up the sample space and are in which we can select two valves from 12 is 12 2 all equally likely to be selected. Hence we can assign a probability of 1/66 to each. The number of pairs
satisfying the condition ‘both are faulty’ is 42 = 6. Consequently


P (both are faulty) =

4 2
12 2

=

1 6 = . 66 11

(ii) This is answered by remembering that P (at least one faulty) = 1 − P (none faulty) . The number of pairs   8(7) 8! 8 = = 28, since this is the number of ways of selecting two valves with neither valve faulty is = 6!2! 2(1) 2 from the eight that work. Consequently,
8 19 2
= . P (at least one faulty) = 1 − 12 33 2 ***Do Exercise A:

Find the number of ways of choosing 4 objects from 6.

11. Read the introduction to random variables in section 13.4.1. Then study section 13.4.2 on discrete random variables, for which the random variable can take only a discrete set of values. Work through Example 13.12. 12. A continuous random variable can take any value in a specified range. Study section 13.4.3 and work through Example 13.13. In this solution you should note that the distribution function is calculated 1 by integrating the density function e−x/2 with respect to x from x = 0 to x. To avoid confusion between 2 the variable of integration x and the upper limit x, J. changes the variable of integration from x to z . The distribution function, FX (x), clearly measures the area under the graph of the density function fX (x), for X ≤ x. 13. Study section 13.4.4 on the properties of density and distribution functions. Work through Example 13.14, noting that the solution uses the expression for FX (x) calculated in Example 13.13. ***Do Exercise 28 on pp.882*** 14. Measures of location and dispersion are considered in section 13.4.6. Read the introductory paragraph and study the subsection on mean, median and mode. Work through Example 13.16. For part (a), observe that             1 1 1 1 1 1 +2 +3 +4 +5 +6 = 3.5 , µX = 1 6 6 6 6 6 6 as stated. The mean in part (b) is calculated in a similar way. 15. Continue section 13.4.6 by studying the subsection on p.884 on variance, standard deviation and quartiles. Look through Example 13.17. –5–

16. Finally read 13.4.7 on expected values. ***Do Exercises 33, 38(a),(b mean only),(c) on p.895***

B:

Work Scheme based on STROUD (FIFTH EDITION)

S. contains much of the material on probability required for this module but some topics are omitted. It is suggested that you work through frames 1–56 of Programme 28 in S, and then go through A: Work scheme based on JAMES (THIRD EDITION) to study the new material, mainly on Venn diagrams and properties of discrete and continuous random variables.

–6–

Specimen Test 24 1.

The sample space S consists of the following set of seven outcomes: S = {copper, sodium, nitrogen, potassium, oxygen, uranium, zinc}, and the events E1 , E2 and E3 are given by E1 = {copper, sodium, zinc}, E2 = {sodium, nitrogen, potassium}, E3 = {oxygen, uranium, zinc}. List the outcomes of the following events: (i) (E1 or E3 ) (ii) (E1 and E2 )

(iii) [(E1 and E3 ) or E2 ] (iv) E2 2.

If P (A) = 0.5, P (B) = 0.3 and P (A or B) = 0.7 , what is P (A and B) ?

3.

One bag contains four white balls and another contains two white balls and two black. One of the bags is selected at random, and a ball is drawn from it which turns out to be white. What is the probability that the remaining balls in the bag are all white?

4.

In how many ways can 4 objects be chosen from 9?

5.

If the probability density function of a random variable X is given by fX (x) =

n

ce−x 0

x>0 otherwise

where c is a constant, find (i) the value of c (ii) the distribution function (iii) P (X > 1).

6.

A discrete random variable can take the values 1, 2, 3, 4, 5 with probabilities Obtain the

7.

(i) mean,

1 1 2 1 1 , , , , . 5 10 5 10 5

(ii) mode.

You are given that fX (x) denotes the probability density function of a continuous random variable which can take any value between 0 and 1. (i) Write down an expression for µ, the mean of the distribution. (ii) Write down an expression for σ 2 , the variance of the distribution.

–7–