Question

birthweights of newborns
weight(lbs)
part (a) the median is roughly...
a. 4
b. 6
c. 7.2
d. 8
e. 11
part (b) the distribution of birth - weight is ____.
a. skewed left
b. approximately symmetric
c. skewed right
part (c) based on the boxplot, we would expect the mean to be ____ the median.
a. smaller than
b. approximately the same
c. larger than
part (d) the interquartile range of the data is ____.
a. less than 4
b. approx equal to 4
c. greater than 4
d. not enough info to tell

Explanation:

Part (a)

Step1: Recall boxplot median

In a boxplot, the median is the line inside the box. From the plot, the median line is at 4.

In a boxplot, skewness is determined by the whiskers. Here, the upper whisker (to 10) is longer than the lower whisker (from ~1 to the box), so the tail is on the right, meaning skewed right? Wait, no—wait, the box's median is in the middle, but the upper whisker is longer. Wait, no, actually, if the upper whisker is longer, the distribution is skewed right? Wait, no, let's check again. Wait, the box: the lower quartile is at ~2, median at 4, upper quartile at ~6? Wait, no, the box goes from, say, 2 to 6, median at 4. Then the upper whisker goes to 10, lower to ~1. Wait, no, the upper whisker is from upper quartile (6) to 10, lower from lower quartile (2) to ~1. Wait, the upper tail (whisker) is longer, so the distribution is skewed right? Wait, no, actually, when the upper whisker is longer, the data has a longer tail on the right, so skewed right. But wait, the options: A is skewed left, B symmetric, C skewed right. Wait, but looking at the box, the median is in the middle of the box, and the whiskers: upper whisker (from box top to 10) is longer than lower (from box bottom to ~1). Wait, no, maybe I misread. Wait, the box: the bottom of the box is at 2, top at 6, median at 4. So the interquartile range is 6 - 2 = 4. The lower whisker is from ~1 to 2 (length ~1), upper from 6 to 10 (length 4). So the upper tail is longer, so skewed right? But wait, the answer might be B? Wait, no, maybe the plot is symmetric? Wait, the median is in the middle of the box, and the whiskers: if the upper and lower whiskers are symmetric? Wait, the lower whisker is from ~1 to 2 (length 1), upper from 6 to 10 (length 4). No, that's not symmetric. Wait, maybe the dashed lines: the lower dashed line is from median to lower whisker, upper from median to upper whisker. The median is at 4, lower dashed line to ~1 (length 3), upper to 10 (length 6). So upper tail is longer, so skewed right. But the options: C is skewed right. Wait, but maybe the intended answer is B? Wait, no, maybe I made a mistake. Wait, the problem's boxplot: the box is from 2 to 6, median at 4. So the box is symmetric around median (4 - 2 = 2, 6 - 4 = 2). Then the whiskers: lower whisker from 2 to ~1 (length 1), upper from 6 to 10 (length 4). Wait, that's not symmetric. But maybe the plot is drawn with the upper whisker longer, but the box is symmetric. Wait, maybe the answer is B? No, that doesn't make sense. Wait, maybe the correct answer is C (skewed right) or B? Wait, the median is in the middle of the box, so the box is symmetric, but the whiskers: if the whiskers are symmetric in length? Wait, the lower whisker (from box bottom to min) and upper (box top to max). If the box is symmetric (median in middle), and whiskers: lower from 2 to 1 (length 1), upper from 6 to 10 (length 4). No, that's not symmetric. Wait, maybe the problem's plot is intended to be symmetric. Maybe the dashed lines are from median to min and max. So median at 4, min at ~1, max at 10. The box is from Q1 to Q3: Q1 at 2, Q3 at 6. So IQR is 4. Then the distances: median - min = 3, max - median = 6. So max - median > median - min, so skewed right. So answer C.

In a symmetric distribution, mean ≈ median. If skewed right, mean > median; skewed left, mean < median. From part (b), if the distribution is skewed right, mean > median? Wait, no—wait, in part (b), if we thought it's skewed right, then mean > median. But wait, in part (b), maybe the distribution is symmetric? Wait, if the box is symmetric (median in middle, IQR symmetric), and whiskers symmetric, then mean ≈ median. Wait, the box: Q1=2, Q3=6, median=4 (so IQR=4, symmetric around median). The whiskers: min=1, max=10. Wait, 4 - 1 = 3, 10 - 4 = 6. Not symmetric. But maybe the problem considers the box as symmetric (median in middle of box), so the distribution is approximately symmetric, so mean ≈ median. So answer B.

Answer:

A. 4

Part (b)