Question

27 33 34 35 37 39 40 41 54 84,75,90,87,99,91,85,88,76,92,94 median: first quartile: third quartile: interquartile range:

Answer 1

For the given data, the median is 85, the first quartile is 37, the third quartile is 90.5, the interquartile range (IQR) is 53.5, and the range is 72.

Step-by-step explanation:

The median in a dataset is the middle value when the values are arranged in numerical order. For the given data 27, 33, 34, 35, 37, 39, 40, 41, 54, 75, 84, 75, 90, 87, 99, 91, 85, 88, 76, 92, 94, after arranging them in ascending order, the median (Q2) is the value positioned at the center of the dataset. Since there are 21 values, the median is the 11th value, which is 85. The first quartile (Q1) is the median of the first half (lower half) of the dataset, which is the value in the 5.5th position, so we average the 5th and 6th values to get 37. Similarly, the third quartile (Q3) is the median of the second half (upper half) of the dataset, which would be the average of the 16th and 17th values, yielding 90.5. The interquartile range (IQR) is the difference between the third quartile and the first quartile, so IQR = Q3 - Q1 = 90.5 - 37 = 53.5. The range of the dataset is the difference between the maximum and minimum value which is 99 - 27 = 72.

Answer 2

From the data set given we shall evaluate the specifics as follows:
27, 33, 34, 35, 37, 39, 40, 41, 54, 84,75,90,87,99,91,85,88,76,92,94

First quartile:
Q1=34

Median
Median=(84+75)/2=79.5

Third Quartile
Q3=76

Interquartile range:
Q3-Q1
=76-34
=42