The variance of observations 112, 116, 120, 125,132 is

`58.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

48.8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

61.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 48.8
Given observations 112, 116, 120, 125, 132
$¯ x = \frac{\sum x_{i}}{n} = \frac{605}{5} = 121$
$\sum (x_{i} - ¯ x)^{2} = 81 + 25 + 1 + 16 + 121 = 244$
$∴$ Variance, $σ^{2} = \frac{\sum (x_{i} - ¯ x)^{2}}{n} = \frac{244}{5} = 48.8$

Suggest Corrections

Similar questions

Two patterns of whole numbers are as follows:
120, 116, 112, A, 104, …
45, 47, 49, B, 53, …

Estimate the value of A $\times$ B

Q. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table.

\begin{matrix} Number of seats & 100 - 104 & 104 - 108 & 108 - 112 & 112 - 116 & 116 - 120 Frequency & 15 & 20 & 32 & 18 & 15 \end{matrix}

Determine the mean number of seats occupied over the flights.

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

Q. The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Find the mean and variance for the following frequency distribution

$\begin{matrix} Classes & 0 - 30 & 30 - 60 & 60 - 90 & 90 - 120 & 120 - 150 & 150 - 180 & 180 - 210 Frequency & 2 & 3 & 5 & 10 & 3 & 5 & 2 \end{matrix}$

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

Join BYJU'S Learning Program

The variance of observations 112, 116, 120, 125,132 is

The correct option is A 48.8Given observations 112, 116, 120, 125, 132¯x=∑xin=6055=121∑(xi−¯x)2=81+25+1+16+121=244∴ Variance, σ2=∑(xi−¯x)2n=2445=48.8

The correct option is A 48.8
Given observations 112, 116, 120, 125, 132
$¯ x = \frac{\sum x_{i}}{n} = \frac{605}{5} = 121$
$\sum (x_{i} - ¯ x)^{2} = 81 + 25 + 1 + 16 + 121 = 244$
$∴$ Variance, $σ^{2} = \frac{\sum (x_{i} - ¯ x)^{2}}{n} = \frac{244}{5} = 48.8$