maths statistics

Slide 1: 

MATHS PROJECT

Slide 2: 

STATISTICS AND PROBABILITY

Slide 3: 

MULTIPLE CHOICE QUESTIONS

Slide 4: 

PROBABILITY:- Q1. A bag contains 6 red, 5 white & 9 black balls. A ball is drawn at random from bag. The probability of drawing a red or white ball is- 11/20 1/4 3/4 7/10 Ans.-11/20

Slide 5: 

Q2. A dice is thrown once. The probability of getting even number is 1/2 1/3 1/4 2/3 Ans.-1/2

Slide 6: 

Q3. One card is drawn from a well shuffled deck of cards. The probability of getting a black queen is 1/26 1/13 3/52 26/52 Ans.-1/26

Slide 7: 

Q4. Two coins are tossed simultaneously. Find the probability of getting at least 1 tail. 3/4 1/4 1/2 1 Ans.-3/4

Slide 8: 

Q5. It is know that a box of 600 ball contains 12 defected balls. One ball is taken out at random from this box. The probability of number of deflected balls is- 49/50 1/50 47/50 1/25 Ans.-49/50

Slide 9: 

STATISTICS:- Q1. In the 2nd game of the 1989 World Series between Oakland and San Francisco 10 players went hitless, 8 players had one hit apiece, and one player had three hits. What were the mean and median number of hits? 11/19, 0 11/19, 1 1, 1 0, 1 Ans.-11/19,0

Slide 10: 

Q2. If MEAN=18, tick the correct value of f. 10 20 30 40 Ans.-20

Slide 11: 

Q3. Suppose the average score on a national test is 500 with a standard deviation of 100. If each score is increased by 25, what are the new mean and standard deviation? 500, 100 500, 125 525, 100 525, 105 525, 125 Ans.-525,105

Slide 12: 

Q4. If Q1 = 20 and Q3 = 30, which of the following must be true? I. The median is 25 II. The mean is between 20 and 30 III. The standard deviation is at most 10. I only II only III only All are true None are true Ans.-All are true.

Slide 13: 

Q5.  What is the median of the following numbers– 83,54,48,60  63.7 63.6 63.5 63.3 Ans:-63.5

Q-Which one of the following is not the measure of central tendency? : 

Q-Which one of the following is not the measure of central tendency? Mean Median Mode Standard deviation Ans. Standard Deviation

Q-The arithmetic mean 1,2,3,…….,n is : 

Q-The arithmetic mean 1,2,3,…….,n is n+1/2 n-1/2 n/2 n/2+1 Ans. n+1/2

Slide 16: 

3 8 13 24 Ans.- 8 Q-If the mean of the following distribution is 2.6, then the value of y is

Q-Mode is- : 

Q-Mode is- Least frequent value Most frequent value Middle most value None of these Ans.- Middle Most Value

Problem : : 

Problem : A coin is tossed three times. What is the probability that it lands on heads exactly one time? (A) 0.125 (B) 0.250 (C) 0.333 (D) 0.375 (E) 0.500 The correct answer is (D). If you toss a coin three times, there are a total of eight possible outcomes. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Of the eight possible outcomes, three have exactly one head. They are: HTT, THT, and TTH. Therefore, the probability that three flips of a coin will produce exactly one head is 3/8 or 0.375.

Slide 19: 

Q. Which of the following statements are true? (Check one) I. Categorical variables are the same as qualitative variables. II. Categorical variables are the same as quantitative variables. III. Quantitative variables can be continuous variables. (A) I only (B) II only (C) III only (D) I and II (E) I and III The correct answer is (E).

Slide 20: 

SHORT QUESTIONS

Q- An unbiased dice is thrown once : 

Q- An unbiased dice is thrown once Write all the possible outcomes. Are the outcomes equally likely? find probability of getting a number < or = 3 find prob. Of getting a number >6

ANSWER : 

ANSWER When an unbiased dice is thrown once,the six possible outcomes are: 1,2,3,4,5,6 Yes, as the dice is unbiased. P (a number _<3) = P(1,2,3)=3/6=1/2 P (a number > 6) = 0/6 =0

Q-In a toss of two coins, find the probability of getting : 

Q-In a toss of two coins, find the probability of getting Both heads Both heads and both tails Ans. 1/4 2/4 = 1/2

Q-Three coins are tossed once.Find the probability of getting : 

Q-Three coins are tossed once.Find the probability of getting Exactly 2 heads Head and tail alternatively Answer: P (exactly 2 heads)= 3/8 P (head and tail alternatively)= 2/8=1/4

One card is drawn from the set of 17 cards numbered 1-17.Find the probability that the number is divisible by : 

One card is drawn from the set of 17 cards numbered 1-17.Find the probability that the number is divisible by 3 7 3 or 7 Both 3 and 7

Answer : 

Answer Numbers divisible by 3 are 3,6,9,12,15=5/17 Numbers divisible by 7 are 7,14=2/17 Numbers divisible by 3 or 7 are 3,6,9,12,15,7,14=7/17 There is no number divisible by both 3 and 7.=0/17=0

Q-A bag contains 10 red, 5 green and 5 blue balls. A ball is drawn randomly from the bag .Find the probability of getting : 

Q-A bag contains 10 red, 5 green and 5 blue balls. A ball is drawn randomly from the bag .Find the probability of getting Blue ball Red ball Red or green ball Green or blue ball

Slide 28: 

Ans. 5/20=1/4 10/20=1/2 10+5/20=3/4 5+5/20=1/2

Slide 29: 

Problem A student goes to the library. The probability that she checks out (a) a work of fiction is 0.40, (b) a work of non-fiction is 0.30, , and (c) both fiction and non-fiction is 0.20. What is the probability that the student checks out a work of fiction, non-fiction, or both?

Slide 30: 

Solution Let F = the event that the student checks out fiction; and let N = the event that the student checks out non-fiction. When either or both of these events occur, we say that the union of F and N has occurred. This problem requires us to find the probability of the union of F and N. To find that probability, we use the rule of additation: P(F ∪ N) = P(F) + P(N) - P(F ∩ N) P(F ∪ N) = 0.40 + 0.30 - 0.20 = 0.50

Slide 31: 

Problem 1 A card is drawn randomly from a deck of ordinary playing cards. You win $10 if the card is a spade or an ace. What is the probability that you will win the game?

Slide 32: 

Solution Let S = the event that the card is a spade; and let A = the event that the card is an ace. You will win the game if either of these two events occur. Therefore, this problem requires you to find the probability of the union of two events. We know the following: There are 52 cards in the deck. There are 13 spades, so P(S) = 13/52. There are 4 aces, so P(A) = 4/52. There is 1 ace that is also a spade, so P(S ∩ A) = 1/52. Therefore, based on the rule of addition: P(S ∪ A) = P(S) + P(A) - P(S ∩ A) P(S ∪ A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13

Slide 33: 

LONG QUESTIONS

For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.In a nationwide survey, suppose 100 boys and 50 girls are sampled. What is the probability that the male sample will have at most three more days of absences than the female sample?(A) 0.025 (B) 0.035 (C) 0.045 (D) 0.055 (E) None of the above : 

For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.In a nationwide survey, suppose 100 boys and 50 girls are sampled. What is the probability that the male sample will have at most three more days of absences than the female sample?(A) 0.025 (B) 0.035 (C) 0.045 (D) 0.055 (E) None of the above Problem 1

ANSWER Problem 1 : 

ANSWER Problem 1 The correct answer is B. The solution involves four steps. Find the mean difference (male absences minus female absences) in the population. μd = μ1 - μ2 = 15 - 10 = 5 Find the standard deviation of the difference. σd = sqrt( σ12 / n1 + σ22 / n2 ) σd = sqrt(72/100 + 62/50) = sqrt(49/100 + 36/50) = sqrt(0.49 + .72) = sqrt(1.21) = 1.1 Find the z-score that produced when boys have three more days of absences than girls. When boys have three more days of absences, the number of male absences minus female absences is three. And the associated z-score is z = (x - μ)/σ = (3 - 5)/1.1 = -2/1.1 = -1.818 Find the probability. This problem requires us to find the probability that the average number of absences in the boy sample minus the average number of absences in the girl sample is less than 3. To find this probability, we enter the z-score (-1.818) into Stat Trek's Normal Distribution Calculator. We find that the probability of a z-score being -1.818 or less is about 0.035. Therefore, the probability that the difference between samples will be no more than 3 days is 0.035.

Problem 2 : 

Problem 2 A die is rolled, find the probability that an even number is obtained. Let us first write the sample space  S of the experiment. S = {1,2,3,4,5,6}  Let E be the event "an even number is obtained" and write it down. E = {2,4,6}  We now use the formula of the classical probability. P(E) = n(E) / n(S) = 3 / 6 = 1 / 2 Solution to Problem2:

PROBLEM 3 : 

PROBLEM 3 Two coins are tossed, find the probability that two heads are obtained. Note: Each coin has two possible outcomes H (heads) and T (Tails).

SOLUTION PROBLEM 3 : 

SOLUTION PROBLEM 3 The sample space S is given by. S = {(H,T),(H,H),(T,H),(T,T)}  Let E be the event "two heads are obtained". E = {(H,H)}  We use the formula of the classical probability. P(E) = n(E) / n(S) = 1 / 4

Which of these numbers cannot be a probability? a) -0.00001 b) 0.5 c) 1.001 d) 0 e) 1 f) 20% : 

Which of these numbers cannot be a probability? a) -0.00001 b) 0.5 c) 1.001 d) 0 e) 1 f) 20% Problem 4

SOLUTION PROBLEM 4 : 

SOLUTION PROBLEM 4 A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1.

Problem 5 : 

Problem 5 Two dice are rolled, find the probability that the sum is a) equal to 1 b) equal to 4 c) less than 13

SOLUTION PROBLEM 5 : 

SOLUTION PROBLEM 5 a) The sample space S of two dice is shown below. S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)          (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)          (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)          (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)          (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)          (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }  Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, hence P(E) = n(E) / n(S) = 0 / 36 = 0  b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence. P(E) = n(E) / n(S) = 3 / 36 = 1 / 12  c) All possible outcomes, E = S, give a sum less than 13, hence. P(E) = n(E) / n(S) = 36 / 36 = 1

Problem 6 : 

Problem 6 A die is rolled and a coin is tossed, find the probability that the die shows an odd number and the coin shows a head.

SOLUTION PROBLEM 6 : 

SOLUTION PROBLEM 6 The sample space S of the experiment described in question 5 is as follows S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)            (1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}  Let E be the event "the die shows an odd number and the coin shows a head". Event E may be described as followsE={(1,H),(3,H),(5,H)}  The probability P(E) is given by P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

Problem 7 : 

Problem 7 A card is drawn at random from a deck of cards. Find the probability of getting the 3 of diamond.

SOLUTION PROBLEM 7 : 

SOLUTION PROBLEM 7 The sample space S of the experiment in question 7 is shown below  Let E be the event "getting the 3 of diamond". An examination of the sample space shows that there is one "3 of diamond" so that n(E) = 1 and n(S) = 52. Hence the probability of event E occuring is given by P(E) = 1 / 52

Problem 8 : 

Problem 8 A card is drawn at random from a deck of cards. Find the probability of getting a queen

SOLUTION PROBLEM 8 : 

SOLUTION PROBLEM 8 The sample space S of the experiment in question 7 is shwon above (see question 6)  Let E be the event "getting a Queen". An examination of the sample space shows that there are 4 "Queens" so that n(E) = 4 and n(S) = 52. Hence the probability of event E occuring is given by P(E) = 4 / 52 = 1 / 13

Problem 9 : 

Problem 9 A jar contains 3 red marbles, 7 green marbles and 10 white marbles. If a marble is drawn from the jar at random, what is the probability that this marble is white?

SOLUTION PROBLEM 9 : 

SOLUTION PROBLEM 9 We first construct a table of frequencies that gives the marbles color distributions as follows  Colorfrequency red3 green7 white10 We now use the empirical formula of the probability  Frequency for white color P(E)=________________________________________________ Total frequencies in the above table = 10 / 20 = 1 / 2

Problem 10 : 

Problem 10 The blood groups of 200 people is distributed as follows: 50 have type A blood, 65 have B blood type, 70 have Oblood type and 15 have type AB blood. If a person from this group is selected at random, what is the probability that this person has O blood type?

SOLUTION PROBLEM 10 : 

SOLUTION PROBLEM 10 We construct a table of frequencies for the the blood groups as follows  Group Frequency a = 50 B = 65 O = 70 AB = 15 We use the empirical formula of the probability  Frequency for O blood P(E)= ________________________ Total frequencies = 70 / 200 = 0.35

Problem 11 : 

Problem 11 a) A die is rolled, find the probability that the number obtained is greater than 4. b) Two coins are tossed, find the probability that one head only is obtained. c) Two dice are rolled, find the probability that the sum is equal to 5. d) A card is drawn at random from a deck of cards. Find the probability of getting the King of heart.

SOLUTION PROBLEM 11 : 

SOLUTION PROBLEM 11 a) 2 / 6 = 1 / 3 b) 2 / 4 = 1 / 2 c) 4 / 36 = 1 / 9 d) 4 / 52 = 1 / 13

Problem 12 : 

Problem 12 For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6. In a nationwide survey, suppose 100 boys and 50 girls are sampled. What is the probability that the male sample will have at most three more days of absences than the female sample? (A) 0.025 (B) 0.035 (C) 0.045 (D) 0.055 (E) None of the above

SOLUTION PROBLEM 12 : 

SOLUTION PROBLEM 12 The correct answer is B. The solution involves four steps. Find the mean difference (male absences minus female absences) in the population. μd = μ1 - μ2 = 15 - 10 = 5   Find the standard deviation of the difference. σd = sqrt( σ12 / n1 + σ22 / n2 ) σd = sqrt(72/100 + 62/50) = sqrt(49/100 + 36/50) = sqrt(0.49 + .72) = sqrt(1.21) = 1.1   Find the z-score that produced when boys have three more days of absences than girls. When boys have three more days of absences, the number of male absences minus female absences is three. And the associated z-score is z = (x - μ)/σ = (3 - 5)/1.1 = -2/1.1 = -1.818 Continued…

Slide 57: 

Find the probability. This problem requires us to find the probability that the average number of absences in the boy sample minus the average number of absences in the girl sample is less than 3. We find that the probability of a z-score being -1.818 or less is about 0.035. Therefore, the probability that the difference between samples will be no more than 3 days is 0.035. Continued…

Slide 58: 

MADE BY:- GROUP-3 Mohammed Aquib, Parantak Sharma, Peeyush Poddar, Pranjal Maheshwari, Prateek Mago, Priyam Singhal,Pranjal Tyagi,Prateek Kastiya