Find the mean deviation from the mean and from median of the following distribution: Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 No . of students 5 8 15 16 6

Computation of mean distribution from the median: Marks Number of students f i Cumulative Frequency Midpoints x i d i = x i - Median f i d i f i x i x i - x f i x i - 27 0 - 10 5 5 5 23 115 25 22 110 10 - 20 8 13 15 13 104 120 12 96 20 - 30 15 28 25 3 45 375 2 30 30 - 40 16 44 35 7 112 560 8 128 40 - 50 6 50 45 17 102 270 18 108 N = 50 ∑ f i d i = 478 1350 ∑ f i x i - 27 = 472 N = 50 , N 2 = 25 The cumulative frequency just greater than N 2 = 25 is 28 and the corresponding class is 20 - 30 Thus, the median class is 20 - 30 Using formula: ∴ l = 20 , F = 13 , f = 15 , h = 10 Median l + N 2 - F f × h = 20 + 8 = 28 Mean distribuction from the median = ∑ f i d i N = 478 50 = 9 . 56 Mean ( x ¯ ) = ∑ f i x i N = 1350 50 = 27 Mean deviation from the mean = 1 50 × ∑ i = 1 n f i x i - 27 = 472 50 = 9 . 44 Mean deviation from the median and the mean are 9 . 56 and 9 . 44 , respectively.

Calculate the mean, median and standard deviation of the following distribution: Class- interval: 31 - 35 36 - 40 41 - 45 46 - 50 51 - 55 56 - 60 61 - 65 66 - 70 Frequency: 2 3 8 12 16 5 2 3

Class Interval f i Midpoint x i u i = x i - 53 4 u i 2 f i u i f i u i 2 31 - 35 2 33 - 5 25 - 10 50 36 - 40 3 38 - 3 . 75 14 . 06 - 11 . 25 42 . 18 41 - 45 8 43 - 2 . 5 6 . 25 - 20 50 46 - 50 12 48 - 1 . 25 1 . 56 - 15 18 . 72 51 - 55 16 53 0 0 0 7 . 8 56 - 60 5 58 1 . 25 1 . 56 6 . 25 12 . 5 61 - 65 2 63 2 . 5 6 . 25 5 42 . 18 66 - 70 3 68 3 . 75 14 . 06 11 . 25 0 N = 51 ∑ i = 1 n f i u i = - 33 . 75 ∑ i = 1 n f i u i 2 = 223 . 38 X ¯ = a + h ∑ i = 1 n f i u i N = 53 + 4 - 33 . 75 51 = 50 . 35 σ 2 = h 2 ∑ i = 1 n f i u i 2 N - ∑ i = 1 n f i u i N 2 = 16 223 . 38 51 - 1139 . 06 2601 = 63 . 07 σ = 63 . 07 = 7 . 94 f i C F ( Cumulative frequency ) 2 2 3 5 8 13 12 25 16 41 5 46 2 48 3 51 ∑ f i = 51 = N N 2 = 25 . 5 Median class interval is 51 − 55 . L = 50 . 5 F = 25 f = 16 h = 5 Median = L + N 2 - F f × h = 50 . 5 + 25 . 5 - 25 16 × 5 = 50 . 5 + 2 . 5 16 = 50 . 65

Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 3 nor by 4 .

Given: Two dice are thrown together. We know that, probability of occurrence of an event = Number of Desired outcomes Total number of outcomes . Let A denote the event of getting sum of the numbers on the two faces divisible by 3 , B denote the event of getting a sum of the numbers on the two faces divisible by 4 . Total outcomes are ( 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 ) Hence, total number of outcomes = 36 . Favourable outcomes is that sum of the numbers on the two faces in neither divisible by 3 nor by 4 . For getting sum of numbers on the two faces divisible by 3 . Favourable outcomes = ( 1 , 2 ) , ( 2 , 1 ) , ( 1 , 5 ) , ( 5 , 1 ) , ( 3 , 3 ) , ( 2 , 4 ) , ( 4 , 2 ) , ( 3 , 6 ) , ( 6 , 3 ) , ( 4 , 5 ) , ( 5 , 4 ) , ( 6 , 6 ) Favourable number of outcomes = 12 ∴ P ( A ) = 12 36 For getting sum of numbers on the two faces divisible by 4 . Favourable outcomes = ( 2 , 2 ) , ( 1 , 3 ) , ( 3 , 1 ) , ( 2 , 6 ) , ( 6 , 2 ) , ( 4 , 4 ) , ( 3 , 5 ) , ( 5 , 3 ) , ( 6 , 6 ) Favourable number of outcomes = 9 ∴ P ( B ) = 9 36 Probability of getting sum of the numbers on the two faces divisible by 3 and 4 = P ( A and B ) = 1 36 From addition theorem of probability, P ( A or B ) = P ( A ) + P ( B ) - P ( A and B ) Probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4 is P ( A ¯ ∩ B ¯ ) = P A ∪ B ¯ = 1 - P A ∪ B = 1 - 12 36 + 9 36 - 1 36 = 1 - 12 + 9 - 1 36 = 1 - 20 36 = 1 - 5 9 = 4 9 Therefore, the probability of getting the sum of the numbers on the two faces is neither divisible by 3 nor by 4 = 4 9 .

The standard deviation of 20 observations is 5 . If each observation is multiplied by 2 , find the standard deviation and variance of the resulting observations.

Let the observations be x 1 , x 2 , x 3 , . . . . . . . . . , x 20 and x ¯ be their mean. Variance = 5 and n = 20 Variance = 1 n ∑ x i - x 2 Therefore, 1 20 ∑ x i - x 2 = 5 ⇒ ∑ x i - x 2 = 20 5 _____________ 1 If each observation is multiplied by 2 , we get new observations. Let new observations be y 1 , y 2 , y 3 , . . . . . . . . . . . , y 20 . Where y i = 2 x i _________________ 2 We need to find variance of new observations. New variance = 1 n ∑ y i - y 2 Calculating y = 1 n ∑ y i ⇒ y = 1 20 ∑ 2 x i ⇒ y = 2 1 20 ∑ x i ⇒ y = 2 x _____________ 3 Equation 1 ⇒ ∑ x i - x 2 = 20 5 ⇒ ∑ 1 2 y i - 1 2 y 2 = 20 5 [From 2 and 3 ] ⇒ 1 2 2 ∑ y i - y 2 = 20 5 ⇒ ∑ y i - y 2 = 80 5 New variance = 1 n ∑ y i - y 2 = 80 5 × 1 20 = 4 5 Therefore, the required variance is 4 5 .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Statistics and Probability>Exercise>Q 64

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Earn 100

State and prove addition theorem on probability.

Important Questions on Statistics and Probability

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Statistics and Probability>Exercise>Q 3

The standard deviation of the first

10

natural numbers is

HARD

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Statistics and Probability>Exercise>Q 4

Find the mean deviation from the mean and from median of the following distribution:

$\begin{matrix} Marks & 0 - 10 & 10 - 20 & 20 - 30 & 30 - 40 & 40 - 50 \\ No . of students & 5 & 8 & 15 & 16 & 6 \end{matrix}$

EASY

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Statistics and Probability>Exercise>Q 5

The weight of coffee in $70$ jars is shown in the following table:

$\begin{matrix} Weight (in grams) : & 200 - 201 & 201 - 202 & 202 - 203 & 203 - 204 & 204 - 205 & 205 - 206 \\ Frequency : & 13 & 27 & 18 & 10 & 1 & 1 \end{matrix}$

Determine the variance and standard deviation of the above distribution.

HARD

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Statistics and Probability>Exercise>Q 6

Calculate the mean, median and standard deviation of the following distribution:

$\begin{array}{lllllllcclc} \begin{array}{l} Class- \\ interval: \end{array} & 31 - 35 & 36 - 40 & 41 - 45 & 46 - 50 & 51 - 55 & 56 - 60 & 61 - 65 & 66 - 70 \\ Frequency: & 2 & 3 & 8 & 12 & 16 & 5 & 2 & 3 \end{array}$