Q1 = 24
Q3 = 29
IQR = 5
Explanation:
First we have to arrange the numbers in assending order
19, 21, 24, 24, 25, 25, 26, 29, 29, 33
The lower half of the set of number is
19, 21, 24, 24, 25
Q1 is the median of the lower half of the data
So, Q1 = 24
The upper half of the data is
25, 26, 29, 29, 33
Q3 is the median of the upper half of the data
Q3 = 29
IQR = Q3 - Q1 = 29 - 24 = 5
Suppose a biologist studying the mechanical limitations of growth among different species of tulips monitors a...
Suppose a biologist studying the mechanical limitations of growth among different species of tulips monitors a national preserve. He collects data on the heights of 10 different types of tulips in the reserve and rounds each height to the nearest centimeter 25, 21, 26, 24,28,31,29,25, 17,24 Compute the first quartile (01), the third quartile (Os), and the interquartile range (TQR) of the data set. cm 03 28 cm IQR 5 cm
Sa of 10 > Suppose a biologist studying the mechanical limitations of growth among different species of tulips monitors a nation preserve. He collects data on the heights of 10 different types of tulips in the reserve and rounds cach height to the nea centimeter. 25, 21,26, 24, 27,30, 29,25, 16,23 Compute the first quartile (Q1). the third quartile (Os), and the interquartile range (IQR) of the data set. 021 cm 03 28 cm IQR 25 cm
Suppose a biologist studying the mechanical limitations of growth among different species of tulips monitors a national preserve. He collects data on the heights of 10 different types of tulips in the reserve and rounds each height to the nearest centimeter 25.21,26. 24,27.30, 29. 25, 16.23 Compute the first quartile (). the third quartile (0s) and the interquartile range (QR) of the data set. 0i cim