This histogram has no title or axis labels. Suggest what the data could be.

Given data can be formulated as a table: x y 1 - 2 9 2 - 3 22 . 5 3 - 4 35 4 - 5 12 Discrete data is the type of data that has clear spaces between values. Continuous data is data that falls in a continuous sequence. Discrete data contains distinct or separate values. On the other hand, continuous data includes any value within range. So, the given data is continuous data as the values of x are in continuous sequence without any gaps. This data has uniform width, the class width is 1 . So, the data could be a continuos variable like heights of students in a class and their frequency.

This histogram has no title or axis labels. Suggest what the data could be.

Given data can be formulated as a table: x y 1 - 2 9 2 - 3 22 . 5 3 - 4 35 4 - 5 12 The given data is continuous data as the values of x are in continuous sequence without any gaps. This data has uniform width, the class width is 1 . An outlier is a data point that is far away from the rest of the dataset. Outliers do not have a formal definition, but are easy to determine by looking at histogram. Therefore, we have no outliers in this data as it is evenly distributed.

The histogram shows the number of hours that Macey spent doing homework each day in June. Use the information in the frequency histogram to construct a grouped frequency table.

Given histogram shows the number of hours that Macey spent doing homework each day in June. Time in hours she spent in a day is plotted on x -axis let it be t and frequency is plotted on y -axis. So, the grouped frequency table is Time (hours) Frequency 1 . 5 ≤ t < 2 . 5 5 2 . 5 ≤ t < 3 . 5 6 3 . 5 ≤ t < 4 . 5 8 4 . 5 ≤ t < 5 . 5 7 5 . 5 ≤ t < 6 . 5 4

The histogram shows the number of hours that Macey spent doing homework each day in June. Use your frequency table to calculate an estimate for the mean time that Macey spent doing homework.

Given histogram shows the number of hours that Macey spent doing homework each day in June. Time in hours she spent in a day is plotted on x -axis let it be t and frequency is plotted on y -axis. For finding estimated mean time for this grouped data we make a column of midpoint of each class interval. So, the grouped frequency table is Time (hours) Frequency f Midpoint of time interval ( x ) 1 . 5 ≤ t < 2 . 5 5 1 . 5 + 2 . 5 2 = 2 2 . 5 ≤ t < 3 . 5 6 2 . 5 + 3 . 5 2 = 3 3 . 5 ≤ t < 4 . 5 8 3 . 5 + 4 . 5 2 = 4 4 . 5 ≤ t < 5 . 5 7 4 . 5 + 5 . 5 2 = 5 5 . 5 ≤ t < 6 . 5 4 5 . 5 + 6 . 5 2 = 6 Now, the mean time is = ∑ f x ∑ f = 2 × 5 + 6 × 3 + 8 × 4 + 7 × 5 + 4 × 6 5 + 6 + 8 + 7 + 4 = 119 30 = 3 . 96 So, the estimated mean time is 3 . 96 hours or 3 hours and 58 minutes.

The histogram shows the lengths of 26 black rats. Write down the modal class.

The histogram shows the lengths of 26 black rats. The modal class is the class interval that is lengths interval with the highest frequency. So, here we have highest frequency of 9 in class 36 - 40 So, 36 - 40 is the modal class of given data