data analysis and probability interpreting a box - and - whisker plot: problem type 2\n(a) describing the…

data analysis and probability interpreting a box - and - whisker plot: problem type 2\n(a) describing the numbers of plants omar sold this past spring\nfor the numbers of plants sold, we see that the first quartile is 60.\nthe first quartile separates the lower 25% of the data values from the upper 75%.\nthis means that the number of plants sold was below 60 for 25% of the days and above 60 for 75% of the days.\nso here is the correct description.\nthe number of plants sold was above 60 on more days than it was below 60.\n(b) finding the interval that gives the middle 50% of the data\nthe data between the first quartile and the third quartile give the middle 50% of the data.\nwe see that the first quartile is 60, and the third quartile is 64.\nso the interval 60 to 64 gives the middle 50% of the data.\n(c) finding the difference of the highest number and the lowest number of plants omar sold in a day (the\nthe highest number of plants sold is 66. the lowest number of plants sold is 54.\nso here is the difference were asked to find,\n66 - 54 = 12\nanswer for explanation\n(a) the number of plants sold was below 60 on more days than it was above 60.\nthe number of plants sold was above 60 on more days than it was below 60.\nthe number of plants sold was below 60 for the same number of days that it was above 60.\n(b) 60 to 64 60 to 63 63 to 64 54 to 63\n(c) 12

data analysis and probability interpreting a box - and - whisker plot: problem type 2\n(a) describing the numbers of plants omar sold this past spring\nfor the numbers of plants sold, we see that the first quartile is 60.\nthe first quartile separates the lower 25% of the data values from the upper 75%.\nthis means that the number of plants sold was below 60 for 25% of the days and above 60 for 75% of the days.\nso here is the correct description.\nthe number of plants sold was above 60 on more days than it was below 60.\n(b) finding the interval that gives the middle 50% of the data\nthe data between the first quartile and the third quartile give the middle 50% of the data.\nwe see that the first quartile is 60, and the third quartile is 64.\nso the interval 60 to 64 gives the middle 50% of the data.\n(c) finding the difference of the highest number and the lowest number of plants omar sold in a day (the\nthe highest number of plants sold is 66. the lowest number of plants sold is 54.\nso here is the difference were asked to find,\n66 - 54 = 12\nanswer for explanation\n(a) the number of plants sold was below 60 on more days than it was above 60.\nthe number of plants sold was above 60 on more days than it was below 60.\nthe number of plants sold was below 60 for the same number of days that it was above 60.\n(b) 60 to 64 60 to 63 63 to 64 54 to 63\n(c) 12

Answer

Explanation:

Step1: Analyze first - quartile concept

The first quartile ($Q_1 = 60$) separates the lower 25% of data from the upper 75%. So, the number of plants sold was below 60 for 25% of the days and above 60 for 75% of the days. Thus, the number of plants sold was above 60 on more days than it was below 60.

Step2: Determine middle 50% interval

The middle 50% of the data is between the first quartile ($Q_1 = 60$) and the third quartile ($Q_3=64$). So the interval is 60 to 64.

Step3: Calculate range

The highest number of plants sold is 66 and the lowest is 54. The difference (range) is calculated as $66 - 54=12$.

Answer:

(a) The number of plants sold was above 60 on more days than it was below 60. (b) 60 to 64 (c) 12