Question:

The mean and variance of eight observation are 9 and 9.25, respectively. If six of the observation are 6,7,10,12,12 and 13, find the remaining two observations.

Answer:

Step 1: Find the sum of the six observations. 6 + 7 + 10 + 12 + 12 + 13 = 60

Step 2: Find the sum of all eight observations. Sum of six observations (60) + sum of remaining two observations = 8 × mean (8 × 9)

Step 3: Solve for the sum of the remaining two observations. 8 × 9 - 60 = 24

Step 4: Find the variance of the eight observations. Variance = 9.25

Step 5: Find the variance of the six observations. Variance of six observations = (6 + 7 + 10 + 12 + 12 + 13)2 / 6 - (60)2 / 6 = (780) / 6 - (3600) / 6 = 130 - 600 = 470

Step 6: Find the variance of the remaining two observations. Variance of eight observations (9.25) - variance of six observations (470) = variance of remaining two observations (455.25)

Step 7: Find the mean of the remaining two observations. Mean of remaining two observations = √(variance of remaining two observations / 2) = √(455.25 / 2) = 21.25

Step 8: Find the remaining two observations. Remaining two observations = mean of remaining two observations (21.25) ± √(variance of remaining two observations / 2) = 21.25 ± √(455.25 / 2) = 21.25 ± 12.5 = 8.75 and 33.75

Question:

The mean and standard deviation of marks obtained by 50 students of a class in three subjects Mathematics, Physics and chemistry are given below : Subject Mathematics Physics Chemistry Mean 42 32 40.9 Standard deviation 12 15 20 Which of the three subjects shows the highest variability in marks and which shows the lowest ?

Answer:

Step 1: Examine the given data. The mean and standard deviation of marks obtained by 50 students of a class in three subjects Mathematics, Physics and Chemistry are given.

Step 2: Calculate the variability in marks for each subject. Variability can be calculated by taking the ratio of standard deviation to the mean.

Step 3: The variability in marks for Mathematics = 12/42 = 0.2857, for Physics = 15/32 = 0.46875 and for Chemistry = 20/40.9 = 0.4912.

Step 4: Compare the variabilities. The subject with the highest variability in marks is Chemistry and the subject with the lowest variability in marks is Mathematics.

Question:

The mean and variance of 7 observation are 8 and 16 respectively. If five of the observation are 2,4,10,12 and 14. Find the remaining two observations.

Answer:

Step 1: Calculate the sum of the five given observations. Answer: 2 + 4 + 10 + 12 + 14 = 42

Step 2: Calculate the sum of all seven observations. Answer: Sum of seven observations = Mean x Number of observations = 8 x 7 = 56

Step 3: Find the sum of the remaining two observations. Answer: Sum of remaining two observations = Sum of seven observations - Sum of five given observations = 56 - 42 = 14

Step 4: Find the two remaining observations. Answer: 14 / 2 = 7 Therefore, the remaining two observations are 7 and 7.

Question:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect which were recorded as 21,21 and 18. Find the mean and standard deviation if the incorrect observation are omitted

Answer:

Step 1: Calculate the mean of the 100 observations. Mean = 20

Step 2: Calculate the standard deviation of the 100 observations. Standard Deviation = 3

Step 3: Calculate the sum of the three incorrect observations. Sum = 21 + 21 + 18 = 60

Step 4: Calculate the new mean by subtracting the sum of the incorrect observations from the total sum of the 100 observations. New Mean = (100 × 20) – 60 = 1400 – 60 = 1340

Step 5: Calculate the new standard deviation by subtracting the sum of the incorrect observations from the total sum of the 100 observations. New Standard Deviation = (100 × 3) – 60 = 300 – 60 = 240

Step 6: Calculate the new mean and standard deviation by dividing the new sums by the number of observations after omitting the incorrect observations. New Mean = 1340 / 97 = 13.8 New Standard Deviation = 240 / 97 = 2.5

Question:

Given that Xˉ is the mean and σ2 is the variance of n observations X1​,X2​…Xn​. Prove that the mean and variance of the observations aX1​,aX2​,aX3​….aXn​ are axˉ and a2σ2 respectively (a=0).

Answer:

Proof:

Step 1: Let X1, X2, X3, …, Xn be n observations with mean X¯ and variance σ2.

Step 2: Let aX1, aX2, aX3, …, aXn be the same observations multiplied by a constant a.

Step 3: The mean of aX1, aX2, aX3, …, aXn is given by

mean = (aX1 + aX2 + aX3 + … + aXn) / n

Step 4: Substituting the values of aX1, aX2, aX3, …, aXn, we get

mean = (aX1 + aX2 + aX3 + … + aXn) / n

= (a(X1 + X2 + X3 + … + Xn)) / n

= a(X1 + X2 + X3 + … + Xn) / n

= a(X¯)

Step 5: Thus, the mean of the observations aX1, aX2, aX3, …, aXn is ax¯.

Step 6: The variance of aX1, aX2, aX3, …, aXn is given by

variance = (a2X12 + a2X22 + a2X32 + … + a2Xn2) / n - (ax¯)2

Step 7: Substituting the values of aX1, aX2, aX3, …, aXn, we get

variance = (a2X12 + a2X22 + a2X32 + … + a2Xn2) / n - (ax¯)2

= (a2(X12 + X22 + X32 + … + Xn2)) / n - (a2(X¯)2

= a2(X12 + X22 + X32 + … + Xn2) / n - a2(X¯)2

= a2σ2

Step 8: Thus, the variance of the observations aX1, aX2, aX3, …, aXn is a2σ2.

Step 9: Therefore, the mean and variance of the observations aX1, aX2, aX3, …, aXn are ax¯ and a2σ2 respectively (a ≠ 0).

Question:

The mean and standard deviation of 20 observation are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases : (i) If wrong item is omitted (ii) If it is replaced by 12

Answer:

(i) If wrong item is omitted

Mean = (10 + 11 + 10 + 10 + 11 + 10 + 11 + 11 + 10 + 11 + 10 + 10 + 11 + 10 + 11 + 11 + 10 + 11 + 10 + 11) / 19 = 10.53

Standard Deviation = √[( (10 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 + (10 - 10.53)2 + (11 - 10.53)2 ) / 18 ] = 0.9

(ii) If it is replaced by 12

Mean = (12 + 11 + 10 + 10 + 11 + 10 + 11 + 11 + 10 + 11 + 10 + 10 + 11 + 10 + 11 + 11 + 10 + 11 + 10 + 11) / 20 = 10.6

Standard Deviation = √[( (12 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 + (10 - 10.6)2 + (11 - 10.6)2 ) / 19 ] = 0.8

Question:

The mean and standard deviation of six observation are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Answer:

1. Mean of the six observations = 8
2. Standard deviation of the six observations = 4
3. Each observation is multiplied by 3
4. New mean of the resulting observations = 8 x 3 = 24
5. New standard deviation of the resulting observations = 4 x 3 = 12