Chapter 10.3, Problem 5PPS

To determine

To calculate minimum, lower quartile, median, upper quartile and maximum

Answer to Problem 5PPS

Median is 73, lower quartile is 66, upper quartile is 82, minimum is 62and maximum is 99

Explanation of Solution

Given information:

City art museum’s annual gala event attendance for last 9 years,

68, 99, 73, 65, 67, 62, 80, 81, 83.

Formula used:

Median is,

  ${(\frac{n+1}{2})}^{th}term$ , where n is frequency

Lower quartile is,

  $Q1=(n+1)\times \frac{1}{4}$

Upper quartile is,

  $Q3=(n+1)\times \frac{3}{4}$

Minimum is the smallest value of the data set.

Maximum is the largest value of the data set.

Calculation:

Arranging the data in an ascending order,

62, 65, 67, 68, 73, 80, 81, 83, 99

For median,

  $n=9$

Substituting the values in median formula,

  $M={(\frac{9+1}{2})}^{th}term$

So,

  $M={\frac{10}{2}}^{th}$

On solving,

  $M=5^{th}term$

Substituting the value.

  $M=73$

Hence, median is 73

For lower quartile,

Substituting the values in lower quartile formula,

  $Q1=(9+1)\times \frac{1}{4}$

So,

  $Q1=10\times \frac{1}{4}$

On solving,

  $Q1=2.5$

Calculating the average of 2^nd and 3^rd observation. Substituting the value of 2^nd and 3^rd observation,

  $Q1=\frac{65+67}{2}$

So,

  $Q1=\frac{132}{2}$

On solving,

  $Q1=66$

Hence, Lower quartile is 66

For upper quartile,

Substituting the values in lower quartile formula,

  $Q3=(9+1)\times \frac{3}{4}$

So,

  $Q3=10\times \frac{3}{4}$

On solving,

  $Q3=7.5$

Hence, substituting the value of 7^th and 8^th observations,

  $Q3=\frac{81+83}{2}$ ,

So,

  $Q3=\frac{164}{2}$

On solving,

  $Q3=82$

Hence, Upper quartile is 82

Minimum is the smallest value of the data set. Substituting the value,

Hence, minimum is 62

Maximum is the largest value of the data set. Substituting the value,

Hence, maximum is 99.