Question

range = maximum - minimum = 23 - 11 = 12
result:
range = 12
q2): a bank branch manager was interested in the waiting times of customers. for this he carried out a survey. a random sample of 12 customers was selected and yielded the following. 5.90, 9.66, 5.79, 8.02, 8.73, 8.01, 10.49, 8.35, 6.68, 6.64, 5.47, 9.91
given: no: of customers = 12
data: 5.90, 9.66, 5.79, 8.02, 8.73, 8.01, 10.49, 8.35, 6.68, 6.64, 5.47, 9.91
required:
(i) average =? (ii) median =? (iii) standard deviation =?
solution:
(i) average:
we know that \\( \overline { x } = \frac { \sum x } { n } \\)
\\( \overline { x } = 5.90 + 9.66 + 5.79 + 8.02 + 8.73 + 8.01 + 10.49 + 8.35 + 6.68 + 5.64 + 5.47 + 9.91 \\)
\\( \overline { x } = \frac { 92.64 } { 12 } \\)
\\( \overline { x } = 7.72 \\)
(ii) median:
arrange the given data in ascending order:
5.47, 5.64, 5.79, 5.90, 6.68, 8.01, 8.02, 8.35, 8.73, 9.66, 9.91, 10.49
here n = 12 (even)
median = \\( \frac { 8.01 + 8.02 } { 2 } \\)
median = 8.01

Explanation:

Step1: Calculate the average

We know that the formula for the average (mean) is \(\bar{x}=\frac{\sum x}{n}\).
First, sum up all the data values: \(5.90 + 9.66+5.79 + 8.02+8.73+8.01+10.49+8.35+6.68+6.64+5.47+9.91 = 92.64\)
Then, divide by the number of data points \(n = 12\). So, \(\bar{x}=\frac{92.64}{12}=7.72\)

Step2: Calculate the median

Arrange the data in ascending order: \(5.47,5.64,5.79,5.90,6.68,8.01,8.02,8.35,8.73,9.66,9.91,10.49\)
Since \(n = 12\) (even), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th values.
The \(\frac{n}{2}=6\)th value is \(8.01\) and the \((\frac{n}{2}+1) = 7\)th value is \(8.02\)
Median\(=\frac{8.01 + 8.02}{2}=\frac{16.03}{2}=8.015\approx8.02\)

Step3: Calculate the standard deviation

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n - 1}}\)
First, calculate \((x_i-\bar{x})^2\) for each \(x_i\):
\((5.90 - 7.72)^2=(-1.82)^2 = 3.3124\)
\((9.66-7.72)^2=(1.94)^2 = 3.7636\)
\((5.79 - 7.72)^2=(-1.93)^2 = 3.7249\)
\((8.02-7.72)^2=(0.3)^2 = 0.09\)
\((8.73-7.72)^2=(1.01)^2 = 1.0201\)
\((8.01-7.72)^2=(0.29)^2 = 0.0841\)
\((10.49-7.72)^2=(2.77)^2 = 7.6729\)
\((8.35-7.72)^2=(0.63)^2 = 0.3969\)
\((6.68-7.72)^2=(-1.04)^2 = 1.0816\)
\((6.64-7.72)^2=(-1.08)^2 = 1.1664\)
\((5.47-7.72)^2=(-2.25)^2 = 5.0625\)
\((9.91-7.72)^2=(2.19)^2 = 4.7961\)

Sum of \((x_i-\bar{x})^2=3.3124+3.7636 + 3.7249+0.09+1.0201+0.0841+7.6729+0.3969+1.0816+1.1664+5.0625+4.7961=32.1724\)
Then, \(s=\sqrt{\frac{32.1724}{12-1}}=\sqrt{\frac{32.1724}{11}}\approx\sqrt{2.9248}\approx1.71\)

Answer:

(i) The average is \(7.72\)
(ii) The median is approximately \(8.02\)
(iii) The standard deviation is approximately \(1.71\)