problem solving in statistics and probability

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Course info, instructors.

• Dr. Jeremy Orloff
• Dr. Jennifer French Kamrin

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• Mathematics

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• Discrete Mathematics
• Probability and Statistics

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Introduction to probability and statistics, problem sets with solutions.

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Teach yourself statistics

How to Solve Probability Problems

You can solve many simple probability problems just by knowing two simple rules:

• The probability of any sample point can range from 0 to 1.
• The sum of probabilities of all sample points in a sample space is equal to 1.

The following sample problems show how to apply these rules to find (1) the probability of a sample point and (2) the probability of an event.

Probability of a Sample Point

The probability of a sample point is a measure of the likelihood that the sample point will occur.

Example 1 Suppose we conduct a simple statistical experiment . We flip a coin one time. The coin flip can have one of two equally-likely outcomes - heads or tails. Together, these outcomes represent the sample space of our experiment. Individually, each outcome represents a sample point in the sample space. What is the probability of each sample point?

Solution: The sum of probabilities of all the sample points must equal 1. And the probability of getting a head is equal to the probability of getting a tail. Therefore, the probability of each sample point (heads or tails) must be equal to 1/2.

Example 2 Let's repeat the experiment of Example 1, with a die instead of a coin. If we toss a fair die, what is the probability of each sample point?

Solution: For this experiment, the sample space consists of six sample points: {1, 2, 3, 4, 5, 6}. Each sample point has equal probability. And the sum of probabilities of all the sample points must equal 1. Therefore, the probability of each sample point must be equal to 1/6.

Probability of an Event

The probability of an event is a measure of the likelihood that the event will occur. By convention, statisticians have agreed on the following rules.

• The probability of any event can range from 0 to 1.
• The probability of event A is the sum of the probabilities of all the sample points in event A.
• The probability of event A is denoted by P(A).

Thus, if event A were very unlikely to occur, then P(A) would be close to 0. And if event A were very likely to occur, then P(A) would be close to 1.

Example 1 Suppose we draw a card from a deck of playing cards. What is the probability that we draw a spade?

Solution: The sample space of this experiment consists of 52 cards, and the probability of each sample point is 1/52. Since there are 13 spades in the deck, the probability of drawing a spade is

P(Spade) = (13)(1/52) = 1/4

Example 2 Suppose a coin is flipped 3 times. What is the probability of getting two tails and one head?

Solution: For this experiment, the sample space consists of 8 sample points.

S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}

Each sample point is equally likely to occur, so the probability of getting any particular sample point is 1/8. The event "getting two tails and one head" consists of the following subset of the sample space.

A = {TTH, THT, HTT}

The probability of Event A is the sum of the probabilities of the sample points in A. Therefore,

P(A) = 1/8 + 1/8 + 1/8 = 3/8

Step-by-Step Statistics Solutions

Get help on your statistics homework with our easy-to-use statistics calculators.

Here, you will find all the help you need to be successful in your statistics class. Check out our statistics calculators to get step-by-step solutions to almost any statistics problem. Choose from topics such as numerical summary, confidence interval, hypothesis testing, simple regression and more.

Statistics Calculators

Tables and graphs, numerical summaries, basic probability, discrete distributions, continuous distributions, sampling distributions, confidence intervals, hypothesis testing, two populations, population variance, goodness of fit, analysis of variance, simple regression, multiple regression, time series analysis.

Statistics and Probability Problems with Solutions

Problems on statistics and probability are presented. The solutions to these problems are at the bottom of the page.

• Given the data set 4 , 10 , 7 , 7 , 6 , 9 , 3 , 8 , 9 Find a) the mode, b) the median, c) the mean, d) the sample standard deviation. e) If we replace the data value 6 in the data set above by 24, will the standard deviation increase, decrease or stay the same?
• Find x and y so that the ordered data set has a mean of 42 and a median of 35. 17 , 22 , 26 , 29 , 34 , x , 42 , 67 , 70 , y
• Given the data set 62 , 65 , 68 , 70 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101, find a) the median, b) the first quartile, c) the third quartile, c) the interquartile range (IQR).
• The exam grades of 7 students are given below. 70 , 66 , 72 , 96 , 46 , 90 , 50 Find a) the mean b) the sample standard deviation
• Twenty four people had a blood test and the results are shown below. A , B , B , AB , AB , B , O , O , AB , O , B , A AB , A , O , O , AB , B , O , A , AB , O , B , A a) Construct a frequency distribution for the data. b) If a person is selected randomly from the group of twenty four people, what is the probability that his/her blood type is not O?
• When a die is rolled and a coin (with Heads and Tails) is tossed, find the probability of obtaining a) Tails and an even number, b) a number greater 3, c) Heads or an odd number,
• A box contains red and green balls. The number of green balls is 1/3 the number of red balls. If a ball is taken randomly from the box, what is the probability that the ball is red?
• A committee of 6 people is to be formed from a group of 20 people. The committee has to have the number of women double that of the men. In how many ways can this committee be formed if there are 12 men?
• A student's marks in five tests are 36%, 78%, 67%, 88% and 98%. The weights for the five tests are 1, 2, 2, 3, 3 respectively. Find the weighted mean ? of the five tests.
• In a group of 40 people, 10 are healthy and every person the of the remaining 30 has either high blood pressure, a high level of cholesterol or both. If 15 have high blood pressure and 25 have high level of cholesterol, a) how many people have high blood pressure and a high level of cholesterol? If a person is selected randomly from this group, what is the probability that he/she b) has high blood pressure (event A)? c) has high level of cholesterol(event B)? d) has high blood pressure and high level of cholesterol (event A and B)? e) has either high blood pressure or high level of cholesterol (event A or B)? f) Use the above to check the probability formula: P(A or B) = P(A) + P(B) - P(A and B).
• A committee of 5 people is to be formed randomly from a group of 10 women and 6 men. Find the probability that the committee has a) 3 women and 2 men. a) 4 women and 1 men. b) 5 women. c) at least 3 women.
• In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that a) exactly 5 have access to the internet. b) at least 6 students have access to the internet.
• The grades of a group of 1000 students in an exam are normally distributed with a mean of 70 and a standard deviation of 10. A student from this group is selected randomly. a) Find the probability that his/her grade is greater than 80. b) Find the probability that his/her grade is less than 50. c) Find the probability that his/her grade is between 50 and 80. d) Approximately, how many students have grades greater than 80?

Solutions to the above Problems

• The given data set has 2 modes: 7 and 9
• order data : 3 , 4 , 6 , 7 , 7 , 8 , 9 , 9 , 10 : median = 7
• (mean) : m = (3+4+6+7+7+8+9+9+10) / 9 = 7
• The standard deviation will increase since 24 is further from away from the other data values than 6.
• x = 36 , y = 77
• median = 75
• first quartile = 69
• third quartile = 81
• interquartile range = 81 - 69 = 12
• sample standard deviation ? 18.6 (rounded to 1 decimal place)
• 1 - (7/24) = 17/24 ? 0.71 (rounded to 2 decimal places)
• ? = ? x P(X = x) = 0×0.24 + 1×0.38 + 2×0.20 + 3×0.13 + 4×0.05 = 1.37
• 1) using definition ? = √[ ? (x - ?) 2 P(X = x) ] = √[ (0-1.37) 2 ×0.24 + (1-1.37) 2 ×0.38 + (2-1.37) 2 ×0.2 + (3-1.37) 2 ×0.13 + (4-1.37) 2 ×0.05 ] ? 1.13 (rounded to 2 decimal places) 2) using computing formula ? = √[ ? x 2 P(X = x) - ? 2 ] = √[ 0 2 ×0.24 + 1 2 ×0.38 + 2 2 ×0.2 + 3 2 ×0.13 + 4 2 ×0.05 - 1.37 2 ] ? 1.13 (rounded to 2 decimal places)
• If there 12 men, then there are 20 - 12 = 8 women. The committee has six people with the number of women double that of the men, hence the committee has 4 women and 2 men. The number of ways of selecting 4 women from 8 is given by: 8 C 4 = 70. The number of ways of selecting 2 men from 12 is given by: 12 C 2 = 66. The number of selecting 4 women and 2 men to form the committee is given by : 8 C 4 × 12 C 2 = 70 × 66 = 4620
• ? = ? x i × f i / ? f i ? x i × f i = 1×2 + 2×6 + 3×10 + 4×6 + 5×2 + 6×2 = 90 ? f i = 2 + 6 + 10 + 6 + 2 + 2 = 28 ? = 90 / 28 ? 3.21 (rounded to 2 decimal places)
• Let the marks be: x 1 = 36%, x 2 = 78%, x 3 = 67%, x 4 = 88%, x 5 = 98% and the respective weights be: w 1 = 1, w 2 = 2, w 3 = 2, w 4 = 3, w 5 = 3. The weighted mean = ? x i ×w i / ? w i ? x i ×w i = 36% × 1 + 78%×2 + 67%×2 + 88%×3 + 98%×3 = 884% ? w i = 1 + 2 + 2 + 3 + 3 = 11 weighted mean = 884% / 11 = 80%
• a) Let x be the number of people with both high blood pressure and high level of cholesterol. Hence (15 - x) will be the number of people with high blood pressure ONLY and (25 - x) will be the number of people with high level of cholesterol ONLY. We now express the fact that the total number of people with high blood pressure only, with high level of cholesterol only and with both is equal to 30. (15 - x) + (25 - x) + x = 30 solve for x: x = 10 b) 15 have high blood pressure,hence P(A) = 15/40 = 0.375 c) 25 have high level of cholesterol, hence P(B) = 25/40 = 0.625 d) 10 have both,hence P(A and B) = 10/40 = 0.25 e) 30 have either, hence P(A or B) = 30/40 = 0.75 f) P(A) + P(B) - P(A and B) = 0.375 + 0.625 - 0.25 = 0.75 = P(A or B)
• a) In what follows n C r = n! / [ (n - r)!r! ] and is the number of combinations of n objects taken r at the time and P(A) is the probability that even A happens. There are 16 C 5 ways to select 5 people (committee members) out of a total of 16 people (men and women) There are 10 C 3 ways to select 3 women out of 10. There are 6 C 2 ways to select 2 men out of 6. There are 10 C 3 * 6 C 2 ways to select 3 women out of 10 AND 2 men out of 6. P(3 women AND 2 men) = 10 C 3 * 6 C 2 / 16 C 5 = 0.412087 b) similarly: P(4 women AND 1 men) = 10 C 4 * 6 C 1 / 16 C 5 = 0.288461 c) similarly: P(5 women ) = 10 C 5 * 6 C 0 / 16 C 5 = 0.0576923 (in 6 C 0 the 0 is for no men) d) P(at least 3 women) = P(3 women or 4 women or 5 women) since the events "3 women" , "4 women" and "5 women" are all mutually exclusive, then P(at least 3 women) = P(3 women or 4 women or 5 women) = P(3 women) + P(4 women) + P(5 women) ? 0.412087 + 0.288461 + 0.0576923 ? 0.758240
• a) If a pupil is selected at random and asked if he/she has an internet connection at home, the answer would be yes or no and therefore it is a binomial experiment. The probability of the student answering yes is 60% = 0.6. Let X be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that X = 5 is given by the binomial probability formula as follows: P(X = 5) = 8 C 5 (0.6) 5 (1-0.6) 3 = 0.278691 b) P(X ? 6) = P(X = 6 or X = 7 or X = 8) Since all the events X = 6, X = 7 and X = 8 are mutually exclusive, then P(X ? 6) = P(X = 6) + P(x = 7) + P(X = 8) = 8 C 6 (0.6) 6 (1-0.6) 2 + 8 C 7 (0.6) 7 (1-0.6) 1 + 8 C 8 (0.6) 8 (1-0.6) 0 ? 0.315394
• a) x = 80 , z = (80 - 70)/10 = 1 Probablity for grade to be greater than 80 = 1 - 0.8413 = 0.1587 b) x = 50 , z = (50 - 70)/10 = -2 Probablity for grade to be less than 50 = 0.0228 c) The z-scores for x = 50 and x = 80 have already been calculated above. Probablity for grade to be between 50 and 80 = 0.8413 - 0.0228 = 0.8185 d) 0.1587 * 1000 ? 159 (rounded to the nearest unit)
• a) 500 - (170+90+60+50) = 130 tons of steel/iron was recycled. b) 60/500 = 0.12 = 12% of the total recycled was glass.

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Applied Probability and Statistics

Eas-305lr-z1, university at buffalo.

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