The stem-and-leaf plot below gives the test scores for Mrs. Yang's history classes. There were 15 students in fourth period and 20 in fifth period. Use the plot to answer the questions. ​

Answers:

Explanation:

Part (a)

We are given a two-sided stem-and-leaf plot.

The stems are along the middle column. We have two sets of leaves (one set per class).

The stems are 5,6,7,8,9. The stems represent the tens digit of the test score. Each leaf is a units digit of the test score, so it means each leaf represents a different student (there are 15+20 = 35 leaves total)

In the bottom row we have the stem 9 to represent all the scores in the 90s. In fourth period, only one leaf is shown and that is 8. The score 98 is the only score in the 90s for fourth period. There are more 90s scores for fifth period because we have the three leaves 2, 2, and 9. They correspond to the scores 92, 92, and 99. It's possible to repeat leaves.

Fifth period has more scores in the 90s

Part (b)

The data set for the fourth period class is

{51, 53, 56, 57, 59, 61, 66, 66, 70, 71, 71, 77, 77, 86, 98}

To get this data set, read off each stem with their corresponding leaf. Eg: in the first row we have a stem of 5 pair with the right-most leaf of 1 to form the score 51. You should have 15 scores in this set.

If the data set isn't sorted from smallest to largest, then make sure to do so. After doing that, cross off the first and last items to end up with this smaller set (two less items in it from the previous set listed above)

{53, 56, 57, 59, 61, 66, 66, 70, 71, 71, 77, 77, 86}

Repeat the last step. Cross off the first and last items to get

{56, 57, 59, 61, 66, 66, 70, 71, 71, 77, 77}

Repeat again

{57, 59, 61, 66, 66, 70, 71, 71, 77}

and again

{59, 61, 66, 66, 70, 71, 71}

we keep going until we reach either one or two values left

{61, 66, 66, 70, 71}

{66, 66, 70}

{66}

The last item remaining is 66, which is the middle-most item. This is the median of the fourth period class scores.

We follow the same steps to find the median of the fifth period class.

The original data set for the fifth period class is

{51,53,58,60,60,61,61,61,64,66,67,73,78,79,85,87,88,92,92,99}

After crossing off each pair of outer items, to reduce the set one step at a time, you should end up with these two items in the very middle: {66,67}

The median is the midpoint of those values, so it is (66+67)/2 = 66.5

Or another way you could do it is to note there are 20 scores in the fifth period class, so the middle is between slots 10 and 11. The 10 is from 20/2 = 10. The values in slots 10 and 11 are 66 and 67 respectively, so the midpoint here is 66.5

So we found:

We can see that the fifth period class had the higher median score.

Part (c)

Fortunately, this part isn't as lengthy as part (b). To get the range, you subtract the largest and smallest items.

The fifth period has a slightly higher range, so those scores are slightly more spread out. More spread out scores means less consistency.

other answer is correct. i had used the radius of the radius somehow. a website: