In calculating the mean and variance of...

MathematicsMeasures of DispersionJEE Advanced 1979Easy

In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the correct figure of 25. He obtained the mean and variance as 45.0 and 16.0 respectively. Determine the correct mean and variance.

Answer:Correct mean 42.3 ; Correct variance 43.81

Visualized Solution (Hindi)

Given Information

Number of readings $n = 10$
Incorrect Mean $\overset{x}{ˉ}_{o l d} = 45.0$
Incorrect Variance $σ_{o l d}^{2} = 16.0$
Incorrect value used $= 52$
Correct value to be used $= 25$

Calculating Incorrect Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{o l d}

Formula for mean: $\overset{x}{ˉ} = \frac{\sum x}{n}$
Rearranging for sum: $\sum x = n \times \overset{x}{ˉ}$
Incorrect sum $\sum x_{o l d} = 10 \times 45.0 = 450$

Correcting the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{n e w}

Correct sum $\sum x_{n e w} = \sum x_{o l d} - (wrong value) + (correct value)$
$\sum x_{n e w} = 450 - 52 + 25$
$\sum x_{n e w} = 423$

Calculating Correct Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \overset{x}{ˉ}_{n e w}

Correct mean $\overset{x}{ˉ}_{n e w} = \frac{\sum x _{n e w}}{n}$
$\overset{x}{ˉ}_{n e w} = \frac{423}{10}$
$\overset{x}{ˉ}_{n e w} = 42.3$

Formula for Variance

Variance formula: $σ^{2} = \frac{\sum x ^{2}}{n} - (\overset{x}{ˉ})^{2}$
We need to find $\sum x^{2}$ to correct it.

Finding Incorrect Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{o l d}^{2}

$16.0 = \frac{\sum x _{o l d}^{2}}{10} - (45.0)^{2}$
$16.0 = \frac{\sum x _{o l d}^{2}}{10} - 2025$
$\frac{\sum x _{o l d}^{2}}{10} = 16 + 2025 = 2041$
$\sum x_{o l d}^{2} = 20410$

Correcting the Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{n e w}^{2}

Correct sum of squares $\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (52)^{2} + (25)^{2}$
$\sum x_{n e w}^{2} = 20410 - 2704 + 625$
$\sum x_{n e w}^{2} = 18331$

Calculating Correct Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ_{n e w}^{2}

Correct variance $σ_{n e w}^{2} = \frac{\sum x _{n e w}^{2}}{10} - (\overset{x}{ˉ}_{n e w})^{2}$
$σ_{n e w}^{2} = \frac{18331}{10} - (42.3)^{2}$
$σ_{n e w}^{2} = 1833.1 - 1789.29$
$σ_{n e w}^{2} = 43.81$

Summary and Takeaway

Correct Mean $= 42.3$
Correct Variance $= 43.81$
Key Takeaway: Always correct the fundamental sums ( $\sum x$ and $\sum x^{2}$ ) before recalculating statistical measures.

00:00 / 00:00

Conceptually Similar Problems

MathematicsVariance and Standard DeviationJEE Advanced 1982Moderate

View in:English Hindi

The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 marks each. The mean and the variance obtained from this distribution are found to be 40 and 49 respectively. It was later discovered that two observations belonging to the class interval (21-30) were included in the class interval (31-40) by mistake. Find the mean and the variance after correcting the error.

Answer:new mean = 39.5, new variance = 49.25

MathematicsMeasures of DispersionJEE Main 2003Easy

View in:English Hindi

In an experiment with 15 observations on x, the following results were available: $\sum x^{2} = 2830, \sum x = 170$ . One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is

(A)

8.33

(B)

78.00

(C)

188.66

(D)

177.33

Answer:B

MathematicsVariance and Standard DeviationJEE Main 2008Easy

View in:English Hindi

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

(A)

a=0, b=7

(B)

a=5, b=2

(C)

a=1, b=6

(D)

a=3, b=4

Answer:D

MathematicsMeasures of Central Tendency (Mean, Median, Mode)JEE Main 2015Easy

View in:English Hindi

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is:

(A)

15.8

(B)

14.0

(C)

16.8

(D)

16.0

Answer:B

MathematicsMeasures of DispersionJEE Main 2013Easy

View in:English Hindi

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ?

(A)

mean

(B)

median

(C)

mode

(D)

variance

Answer:D

MathematicsVariance and Standard DeviationJEE Main 2012Easy

View in:English Hindi

Let $x_{1}, x_{2}, \dots, x_{n}$ be n observations, and let $\overset{x}{ˉ}$ be their arithmetic mean and $σ^{2}$ be the variance. Statement-1 : Variance of $2 x_{1}, 2 x_{2}, \dots, 2 x_{n}$ is $4 σ^{2}$ Statement-2 : Arithmetic mean $2 x_{1}, 2 x_{2}, \dots, 2 x_{n}$ is $4 \overset{x}{ˉ}$ .

(A)

Statement-1 is false, Statement-2 is true.

(B)

Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

(C)

Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

(D)

Statement-1 is true, statement-2 is false.

Answer:D

MathematicsMeasures of DispersionJEE Main 2014Easy

View in:English Hindi

The variance of first 50 even natural numbers is

(A)

437

(B)

437/4

(C)

833/4

(D)

833

Answer:D

MathematicsMeasures of DispersionJEE Main 2006Easy

View in:English Hindi

Suppose a population A has 100 observations 101, 102, ..., 200 and another population B has 100 observations 151, 152, ..., 250. If $V_{A}$ and $V_{B}$ represent the variances of the two populations, respectively then $\frac{V _{A}}{V _{B}}$ is

(A)

(B)

\frac{9}{4}

(C)

\frac{4}{9}

(D)

\frac{2}{3}

Answer:A

MathematicsBinomial DistributionJEE Main 2003Easy

View in:English Hindi

The mean and variance of a random variable $X$ having binomial distribution are 4 and 2 respectively, then $P (X = 1)$ is

(A)

1/4

(B)

1/32

(C)

1/16

(D)

1/8

Answer:B

MathematicsMean DeviationJEE Main 2009Easy

View in:English Hindi

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to:

(A)

20.0

(B)

10.1

(C)

20.2

(D)

10.0

Answer:B