MAT 211 - Final



November 20, 2011: Basic Probability Rule (Material for final exam Fall 2011)

1. Given [pic] and[pic]. Find[pic].

Answer: [pic]

2. From the given table find [pic]

| |B |C |D |K |Total |

|E |7 |3 |7 |5 |22 |

|F |8 |6 |11 |4 |29 |

|Total |15 |9 |18 |9 |51 |

Answer: [pic]160/493=0.3245436105

3. A family has two children. Using b to stand for boy and g for girl in ordered pairs write the sample space then find the event E that the family has exactly one daughter and the event F the family has at least one daughter. Now identify which one is true

A) E and F are independent (B) E and F are dependent

C) E and F are mutually exclusive that means [pic] (D) None of the given

Answer: (B) Dependent

4. Roll a die twice. Write the sample space. Find the following probabilities:

[pic]

Answer: 5/6

5. Suppose events A and B are independent and that p(A) = 0.4 and p(B) = 0.6. Find[pic].

Answer: 0.24

6. Suppose events A and B are independent and that p(A) = 0.3 and p(B) = 0.6. Find[pic]

Answer: 0.72

7. Given that p(A) = 0.4, p(B) = 0.5 and [pic]. Find [pic].

Answer: 0.70

8. Events [pic], [pic] and [pic] form a partition of the sample space S with probabilities [pic], [pic]and [pic]

If E is an event in S with [pic], [pic] and [pic] then find the probability of [pic]

Answer: 0.47

9. A package, say A, of 24 crocus bulbs contains eight yellow, eight white, and eight purple crocus bulbs. A package, say B, of 24 crocus bulbs contains six yellow, six white and 12 purple crocus bulbs. One of the two packages is selected at random.

a. If three bulbs from this package were planted and all three yielded white flowers, compute the conditional probability that the package A was selected.

b. If the three bulbs yielded one yellow, one white, and one purple flower, compute the conditional probability that the package B was selected.

Solution: Suppose WWW means three white bulbs, then [pic], [pic], also [pic]

a)

[pic]

b)

[pic]

10. How many different license plates can be made using 2 letters followed by 4 digits, if

a. Letters and digits may be repeated. Answer: 6760000

b. Letters may be repeated, but digits are not repeated. Answer: 3407040

c. Neither letters nor digits are repeated. Answer: 3276000

11. License plates in a small state have three letters followed by four numbers. How many possible license plates are possible if the letters have to be different? Repeated numbers are also not allowed. Answer: 78624000

12. A group of musicians consists of 3 percussionists, 4 cellists and 4 violinists. A subgroup needs to be picked consisting of 1 percussionists, 1 cellist and up to 2 violinist. How many such sub groupings are possible? Answer: 204

13. Write the universal set for a two children family: use b for boys and g for girls. Also determine if the set that “the first child a girl” and the set that “the second child is a boy” are mutually exclusive. Write all information in the spaces provided below:

Write universal set U = { Second child is a boy A = {

First child is a girl B = { The intersection: [pic]= {

Conclusion: Not mutually exclusive

Probability Distribution

Random variables of the discrete type

In probability theory, a probability distribution is called discrete if it is characterized by a probability mass function. Thus, the distribution of a random variable X is discrete, and X is then called a discrete random variable, if [pic]

If a random variable is discrete, then the set of all values that it can assume with non-zero probability is finite or countably infinite, because the sum of uncountably many positive real numbers (which is the least upper bound of the set of all finite partial sums) always diverges to infinity.

Given a random experiment with an outcome space S, a function X that assigns to each element s in S one and only one real number X(s) = x is called a random variable, like a function of s. The space of X is the set of real numbers [pic], where [pic] means the element s belongs to S. The probability mass function (pmf) [pic] of a discrete random variable X is a function that satisfies the following properties:

1. [pic]

2. [pic]

3. [pic]

Example 1. Suppose that X has a discrete uniform distribution on [pic]and its pmf is [pic].

As a general case we may write pmf as [pic]

Example 2. Roll a 4 –sided die twice and let X equal the larger of the two outcomes if they are different and common value if they are the same. The outcome space for this experiment is [pic], where we assume that each of these 16 points has probability [pic]. Then [pic] and [pic]. Looking at the pattern one can easily find the pmf

[pic]

Exercise

1. Let the pmf of X be defined by [pic], a) draw a bar graph and a b) probability histogram

2. For each of the following, determine the constant c, so that [pic]satisfies the conditions of being a pmf for a random variable X,

a) [pic] Answer: c = 10

b) [pic] Answer: c = 55

c) [pic] Answer: c = 3

d) [pic] Answer: c = 1/30

e) [pic] Answer: c = n(n+1)/2

Mathematical expectation

In probability theory and statistics, the expected value (or expectation value, or mathematical expectation, or mean, or first moment) of a random variable is the integral of the random variable with respect to its probability measure.

For discrete random variables this is equivalent to the probability-weighted sum of the possible values.

The term "expected value" can be misleading. It must not be confused with the "most probable value." The expected value is in general not a typical value that the random variable can take on. It is often helpful to interpret the expected value of a random variable as the long-run average value of the variable over many independent repetitions of an experiment.

When it exists, mathematical expectation E satisfies the following properties:

a) If c is a constant, [pic]

b) If c is a constant and u is a function, [pic]

c) If [pic] and [pic] are constants and [pic] and [pic] are functions, than [pic]

Example 1. Let X have the pmf [pic]. Find [pic]

Solution: [pic], verify.

Example 2. Let X have the pmf [pic]. Find mean = [pic] and also [pic] and variance [pic] and also standard deviation[pic].

Solution: Mean = [pic]

[pic]

Variance [pic] and [pic]

Example 3. A politician can emphasize jobs or the environment in her election campaign. The voters can be concerned about jobs or the environment. A payoff matrix showing the utility of each possible outcome is shown.

[pic]

The political analysts feel there is a 0.39 chance that the voters will emphasize jobs. Which strategy should the candidate adopt to gain the highest utility

a) Environment b) Jobs

Explain mathematically.

Solution: For the environment the expected value is [pic]

On the other hand for jobs the expected value is [pic]. So the preference will go for a) Environment (because of higher expected value).

Exercise

1. Find mean and standard deviation of the following:

a) [pic] Answer: Mean 15, SD = 7.07

b) [pic] Answer: Mean 5, SD = 0

c) [pic] Answer: Mean 1.67, SD = 0.75

2. Given [pic], determine [pic]Var[pic]and mean=[pic]

Answer: [pic]

[pic]

Bernoulli trials and the Binomial distribution

A Bernoulli experiment is a random experiment, the outcome of which can be classified in but one of two mutually exclusive and exhaustive ways, say, success or failure (life or death, head or tail, 3 or not 3 etc. The pmf of a Bernoulli trail is [pic] and we say that the random variable x has Bernoulli distribution. The mean of Bernoulli trial is given as

[pic], verify.

The variance of Bernoulli trial is

[pic]

Example 1. In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased, the probability of purchasing two winning tickets is [pic]

One may use calculator as follows (TI)

2nd DISTR 0 (binompdf) (8, 0.20, 2) will display 0.29360128

Example 2. In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased, the probability of purchasing at best 6 winning tickets is [pic]

One may use calculator as follows (TI)

2nd DISTR A (binomcdf) (8, 0.20, 6) will display 0.99991552

Example 3. In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased. Find the probability of purchasing at least 6 winning tickets.

Hint. Find [pic] or [pic]

Example 4. A quiz consists of 24 multiple choice questions. Each question has 5 possible answers, only one of which is correct. If you answer the questions completely based on guessing, what is the probability that

a) You will answer exactly 4 wrong?

b) You will answer exactly 4 correctly?

c) You will answer at least 20 correctly?

d) You will answer at most 3 wrong?

e) You will answer at most 3 correctly?

Solution: The probability that you will answer one question wrong is [pic].

a) The probability of answering exactly 4 wrong is a binomial probability of B(24, 0.8, 4), which is [pic], which is almost zero.

If you use TI calculator use binompdf (24, 0.8, 4). Check your calculator using the following code:

2nd DISTR 0 binompdf (24, .8, 4)

b) The probability that you will answer exactly 4 correct is B(24, 0.2, 4) = 0.196

c) At least 20 correct [pic]

= 20 correct + 21 correct + 22 correct + 23 correct + 24 correct = [pic].

It is easy to use calculator with binomcdf as follows:

[pic]

d) At most three wrong: [pic]

e) At most three correct: [pic]

Example 5. A computer manufacturer tests a random sample of 28 computers. The probability that a computer is non defective is 91.3%. What is the probability that:

a) Exactly 7 computers are defective? Answer: 0.006605

b) At least two computers are defective? Answer: 0.7131689

c) At most two computers are defective? Answer: 0.555224

Example 6. A quiz consists of 10 multiple choice questions, each with 4 possible choices. For someone who makes random guesses for all of the questions, find the probability of passing if the minimum passing grade is 90%.

Solution: [pic]

Example 8. A student claims that he has extrasensory perception (ESP). A coin is flipped 25 times, and a student is asked to predict the outcome in advance. He gets 20 out of 25 correct. What was the probability that he would have done at least this well if he had no EPS?

Solution: [pic]

Exercise 1. Toss a fair coin 12 times. How many possible outcomes do you have? What is the probability of getting a) exactly 7 heads, b) at least 7 heads, c) at most 7 heads?

Answer: 4056, a) 19.34% b) 38.72% c) 80.62%

Exercise 2. A student claims that he has extrasensory perception (ESP). A coin is flipped 30 times, and a student is asked to predict the outcome in advance. He gets 25 out of 30 correct. What was the probability that he would have done at least this well if he had no EPS?

Answer: 0.0001625

Exercise 3. A quiz consists of 20 multiple choice questions, each with 5 possible choices. For someone who makes random guesses for all of the questions, find the probability of passing if the minimum passing grade is 80%. Answer: 0.0000000138

Example 4. A computer manufacturer tests a random sample of 30 computers. The probability that a computer is defective is [pic]%. What is the probability that:

a) Exactly 7 computers are defective?

b) At least two computers are defective?

c) At most two computers are defective?

Exercise 5. In the instant lottery with 10% winning tickets, if X is equal to the number of winning tickets among 20 tickets that are purchased, find the probability of purchasing

a) at best 7 winning tickets,

b) at least 7 winning tickets,

c) no more than 6 winning tickets,

d) no less than 6 winning tickets

Exercise 6. The rates of on-time flights for commercial jets are continuously tracked by the U.S Department of transportation. Recently, Southwest Air had the best rate with 80% of its flights arriving on time. A test is conducted by randomly selected 16 Southwest flights and observing whether they arrive on time. Find

a) the probability that exactly 4 flights arrive on time

b) The probability that at least 4 flights arrive on time

c) At best 4 flights arrive on time

Random variable of the continuous type: A random variable is a function X that assigns to each element s in the outcome space S one and only one corresponding real number X(s) = x. The space of X is the set of real numbers [pic]is an interval. In discrete case the S is the set of discrete points. In the continuous case we call the integrable function[pic], a probability density function (pdf) which satisfies the following:

a) [pic] b) [pic]

c) The probability of the event [pic] is [pic]

Probability Distribution Function: A function F is a distribution function of the random variable X iff the following conditions are satisfied:

a) F is non decreasing i.e., [pic], or [pic] for all [pic]

b) F is continuous c) F is normalized i.e., [pic]

Example 1. Evaluate the integral [pic]

Solution: [pic]

Example 2. Show that [pic] is a probability density function.

Solution (Hint): Show that [pic] and [pic]

Example 3. Let Y be a continuous random variable with pdf [pic] and the distribution function is defined by

[pic]

Find mean [pic] (check the integral) and

Variance [pic] (check the integral). Find also the standard deviation [pic].

Example 4. The probability density function of a continuous random variable x is given by the function [pic]. Find its corresponding distribution function, mean, variance and standard deviation, interval of one standard deviation of mean, two standard deviation of mean and three standard deviation of mean.

Solution: The distribution function is the integral of the 1

pdf function over the real line.

Draw the graph of the pdf and notice that F(0) = 0,

0 1 2

F(1) = 1/2 and F(2) = 1. We also notice that distribution function is zero, i.e., [pic] when x < 0.

The distribution function over the interval [pic] is

[pic]

The distribution function over the interval [pic] is

[pic]

The distribution function over the interval [pic] is

[pic]

Thus we have the probability distribution function defined as follows:

[pic]

You can calculate mean, standard deviation and variance. Look at example 3.

Example 5. Show the following function is a probability distribution function

[pic]

Solution: We need to check the following properties:

a) Check that F is non-decreasing.

[pic]

shows that F is not decreasing.

b) For continuity check that [pic] and [pic]

c) F is normalized: [pic]

The function F(x) defined above is probability distribution function.

Exercise:

1. For each of the following functions, i) find the constant c so that f (x) is a pdf of the random variable X, ii) find the distribution function F(x) [pic]and iii) sketch f (x) and F(x), iv) find also [pic] .

a) [pic]

b) [pic]

c) [pic]

d) [pic]

2. Sketch the graph of the following pdf f (x), then find and sketch the probability distribution function F(x) on the real line. Review example 4.

a) [pic]

b) [pic]

c) [pic]

Section 9.4 The Normal Distribution

[pic]

A normal distribution of a random variable X with mean [pic] and variance [pic]is a statistic distribution with probability density function (pdf)

|[pic] |(1) |

on the domain [pic]. While statisticians and mathematicians uniformly use the term "normal distribution" for this distribution, physicists sometimes call it a Gaussian distribution and, because of its curved flaring shape, social scientists refer to it as the "bell curve."

De Moivre developed the normal distribution as an approximation to the binomial distribution, and it was subsequently used by Laplace in 1783 to study measurement errors and by Gauss in 1809 in the analysis of astronomical data (Havil 2003, p. 157).

The normal distribution is an extremely important probability distribution in many fields. It is a family of distributions of the same general form, differing in their location and scale parameters: the mean ("average") and standard deviation ("variability"), respectively. The standard normal distribution is the normal distribution with a mean of zero and a standard deviation of one (the green curve in the plots below). It is often called the bell curve because the graph of its probability density resembles a bell.

[pic]

If a random variable X has this distribution, we write [pic]~ [pic]. If [pic]and [pic], the distribution is called the standard normal distribution and the probability density function reduces to

[pic]

Area under a normal curve:

For a standard normal variate z, the normal distribution has mean zero and standard deviation one with pdf [pic]

The area under the standard normal distribution curve for [pic]. We have now the difficulty to evaluate the integral without having the knowledge of multivariable calculus and polar coordinate form. But this difficulty we can manage using standard values from the table 5 of Normal distribution at page # 423 or using our calculator. Look at example 3.

Important Information: All normal density curves satisfy the following property which is often referred to as empirical rule:

1. 68.26% of the observations fall within 1 standard deviation of mean.

2. 95.44% of the observations fall within 2 standard deviation of mean

3. 99.74% of the observations fall within 3 standard deviation of mean

Note: Within 5 standard deviation of mean we assume 100% data points.

Example 1. Find the mean and standard deviation of the normal distribution whose pdf is given as [pic]

Solution: Compare with the standard formula of pdf for the normal distribution and find that [pic].

Example 2. Write the pdf of a normal distribution with mean 3 and variance 16.

Solution: We have [pic], the pdf of the normal distribution is given as

[pic]

Example 3. Find the area under the normal curve with mean zero and standard deviation one for the standard variate [pic].

Solution: From table 5a:

[pic]

For this value choose row with 1.2 and column 0.04.

0 1.24

Using calculator: [pic]

The calculator code:

2nd DISTR 2 normalcdf (-5, 1.24) =0.8925120

Example 4. Find the area under the normal curve with mean zero and standard deviation one for the standard variate [pic].

Solution: From table 5a:

[pic]

For this value choose row with 1.2 and column 0.04.

0 1.24

Using calculator: [pic]

The calculator code: 2nd DISTR 2 normalcdf ( 1.24, 5) =0.1074875

Example 5. Find the area under the normal curve with mean zero and standard deviation one for [pic].

Solution: From table 5a:

[pic]

-0.12 0 1.24

Using calculator: [pic]

The calculator code:

2nd DISTR 2 normalcdf (-0.12, 1.24) =0.4402707

Example 6. Suppose x is a normally distributed random variable with mean 10.2 and standard deviation 1.5. Find each of the following probabilities.

a) [pic].

b) [pic]

c) [pic]

d) [pic]

e) [pic]

Draw normal curve and show the region bounded by the normal curve and the x values.

Solution: from calculator a) [pic]

try similar way for b), and c),

d) [pic]

Try for e).

6.1 10.2 13.3

Exercise Set

1. The physical fitness of an athlete is often measured by how much oxygen the athlete takes in (which is recorded in millimeters per kilogram, ml/kg). The maximum oxygen uptake for elite athletes has been found to be 80 with a standard deviation 9.2. Assume that distribution is approximately normal.

a) What is the probability that an elite athlete has a maximum oxygen uptake of at least 75 ml/kg? Answer: 70.66%

b) What is the probability that an elite athlete has a maximum oxygen uptake of 65 ml/kg or lower? Answer: 5.15%

c) Consider someone with a maximum oxygen uptake of 26 ml/kg. Is it likely that this person is an elite athlete? Answer: No

2. The combined score of SAT – 1 test are normally distributed with mean of 998 and a standard deviation of 202. If a college includes a minimum score of 800 among its requirements, what percentage of students do not satisfy that requirement? Answer: 16.35%

3. IQ score are normally distributed with mean of 100 and a standard deviation 15. Mensa is an international society that has one – and only one qualification for membership, a score in the top 2 on an IQ test.

a) What IQ score should one have in order to be eligible for Mensa?

Answer: hint: (x-100)/15 = invnorm(0.98), x = 130.81

b) In a typical region of 90,000 people, how many are eligible for Mensa?

Answer: 90,000 (0.02) = 1800

4. Using diaries for many weeks, a study on the lifestyle of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.6 hours of sleep, with a standard deviation of 2.56 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

a) What is the probability that a visually impaired student gets less than 6.1 hours of sleep? Answer: 8.58%

b) What is the probability that a visually impaired student gets between 6.3 and 10.35 hours of sleep? Answer: 51.65%

c) Forty percent of students get less than how many hours of sleep on a typical day? Answer: 8.95 hours

5. Healthy people have body temperatures that are normally distributed with a mean of 98.20 degree Fahrenheit and a standard deviation of 0.62 degree Fahrenheit.

a) If a healthy person is randomly selected, what is the probability that he or she has a body temperature above 98.9 degree Fahrenheit? Answer: 12.94%

b) A hospital wants to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 1% of healthy people to exceed it? Answer: hint: (x-98.2)/.62 = invnorm(0.99), 99.64

6. The heights of a large group of people are assumed to be normally distributed. Their mean height is 68 inches, and the standard deviation is 4 inches. What percent of these people are taller than 73 inches? Answer: 10.56%

7. Suppose a population is normally distributed with a mean of 24.6 and a standard deviation of 1.3. What percent of the data will lie between 25.3 and 26.8? Answer: 24.91%

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