Question

please find the range, sample standard deviation and inter - quartile range (iqr) of the following data set.
29 30 52 53 61 66 66 76 82 96
range = (please enter an exact answer.) standard deviation (s) =
(please show your answer to one decimal place.) iqr =
(please enter an exact answer.)
a new number, 122, is added to the data set above. please find the new range, sample standard deviation and iqr of the new data set.
range =
(please enter an exact answer.) standard deviation =
(please show your answer to one decimal place.) iqr =
(please enter an exact answer.)
which measure of spread is less affected by the addition of the extreme observation?
standard deviation
iqr
range

Explanation:

Step1: Calculate the range of the original data - set

Range is the difference between the maximum and minimum values. The minimum value in the data - set {29, 30, 52, 53, 61, 66, 66, 76, 82, 96} is 29 and the maximum is 96.
$96 - 29=67$

Step2: Calculate the sample standard deviation of the original data - set

First, find the mean $\bar{x}$.
$\bar{x}=\frac{29 + 30+52+53+61+66+66+76+82+96}{10}=\frac{611}{10}=61.1$
Then, calculate the squared differences $(x_i-\bar{x})^2$ for each data point:
$(29 - 61.1)^2=(-32.1)^2 = 1030.41$
$(30 - 61.1)^2=(-31.1)^2 = 967.21$
$(52 - 61.1)^2=(-9.1)^2 = 82.81$
$(53 - 61.1)^2=(-8.1)^2 = 65.61$
$(61 - 61.1)^2=(-0.1)^2 = 0.01$
$(66 - 61.1)^2=(4.9)^2 = 24.01$
$(66 - 61.1)^2=(4.9)^2 = 24.01$
$(76 - 61.1)^2=(14.9)^2 = 222.01$
$(82 - 61.1)^2=(20.9)^2 = 436.81$
$(96 - 61.1)^2=(34.9)^2 = 1218.01$
The sum of squared differences $\sum_{i = 1}^{n}(x_i-\bar{x})^2=1030.41+967.21+82.81+65.61+0.01+24.01+24.01+222.01+436.81+1218.01 = 4070.9$
The sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{4070.9}{9}}\approx21.2$

Step3: Calculate the inter - quartile range (IQR) of the original data - set

First, find the median. Since $n = 10$ (an even number), the median is the average of the 5th and 6th ordered values. Median $M=\frac{61 + 66}{2}=63.5$
The lower half of the data is {29, 30, 52, 53, 61}, and its median (first quartile $Q_1$) is 52.
The upper half of the data is {66, 66, 76, 82, 96}, and its median (third quartile $Q_3$) is 76.
$IQR=Q_3 - Q_1=76 - 52 = 24$

Step4: Calculate the range of the new data - set

The new data - set is {29, 30, 52, 53, 61, 66, 66, 76, 82, 96, 122}. The minimum value is 29 and the maximum is 122.
$122 - 29 = 93$

Step5: Calculate the sample standard deviation of the new data - set

The new mean $\bar{x}=\frac{29+30 + 52+53+61+66+66+76+82+96+122}{11}=\frac{733}{11}\approx66.64$
Calculate the squared differences $(x_i-\bar{x})^2$ for each data point and sum them up. Then, $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$. After calculation, $s\approx27.4$

Step6: Calculate the inter - quartile range (IQR) of the new data - set

The new data - set has $n = 11$ values. The median is the 6th value, which is 66.
The lower half of the data is {29, 30, 52, 53, 61}, and its median (first quartile $Q_1$) is 52.
The upper half of the data is {76, 82, 96, 122}, and its median (third quartile $Q_3$) is 96.
$IQR=Q_3 - Q_1=96 - 52 = 44$

Step7: Determine the measure less affected by the extreme value

The inter - quartile range (IQR) is less affected by extreme values. The range is highly affected as it depends only on the maximum and minimum values. The standard deviation is also affected by extreme values as it takes into account the deviation of each data point from the mean.

Answer:

range (original data - set): 67
standard deviation (original data - set): 21.2
IQR (original data - set): 24
range (new data - set): 93
standard deviation (new data - set): 27.4
IQR (new data - set): 44
Measure less affected by the addition of the extreme observation: IQR