AP Statistics 1.7 Summary Statistics for One Quantitative Variable- Exam Style Questions - FRQs - New Syllabus

(a)
For High School A (\(n = 200\)), the median is the average of the \(100^{\text{th}}\) and \(101^{\text{st}}\) ordered values.

From the histogram, the number of teachers with teaching year in \((1, 7)\) is:
\(46 + 48 = 94 < 100\)
The number with teaching year in \((1, 10)\) is:
\(46 + 48 + 45 = 139 > 100\)
Since there are fewer than \(100\) values below \(7\) but more than \(100\) values below \(10\), both the \(100^{\text{th}}\) and \(101^{\text{st}}\) values fall in the interval \((7, 10)\).
\(\boxed{\text{High School A has median teaching year } = 7}\)

For High School B (\(n = 221\)), the median is the \(111^{\text{th}}\) ordered value.
From the histogram, the number of teachers with teaching year in \((1, 7)\) is:
\(79 + 34 = 113 > 111\)
Since there are more than \(111\) values below \(7\), the \(111^{\text{th}}\) value falls in the interval \((4, 7)\).
\(\boxed{\text{High School B has median teaching year } = 6}\)

Think of the median as a balancing point — you need to find where the “middle person” sits in the ordered list.
For School A, with 94 teachers below year 7 and the median being teacher #100, that teacher has to be somewhere in years 7–9.
For School B, there are already 113 teachers in years 1–6, so the 111th teacher — the median — must be in that same lower group.
The dramatically right-skewed shape of School B also gives a useful intuition: right-skewed distributions pull the mean upward away from the median, and since both schools share the same mean of 8.2, School B (the more skewed one) must have the bigger mean-median gap and therefore the lower median.

(b)
The mean of a combined group is a weighted average — you cannot simply average the two means.
Total teaching years for the original \(200\) teachers: \(\bar{x}_1 \cdot n_1 = 8.2 \times 200 = 1{,}640\)
Total teaching years for the additional \(18\) teachers: \(\bar{x}_2 \cdot n_2 = 2.5 \times 18 = 45\)
Combined mean for all \(218\) teachers: \(\bar{x}_{\text{combined}} = \frac{1{,}640 + 45}{200 + 18} = \frac{1{,}685}{218}\)
\(\boxed{\bar{x}_{\text{combined}} \approx 7.73 \text{ years}}\)
The 18 new teachers have a very low mean of 2.5 years — they are almost all brand-new — so they drag the overall mean down from 8.2 toward 7.73.
Because there are so few of them (only 18 out of 218), the pull is modest but real.
Always weight by group size when combining means; plain-averaging the means (i.e., \((8.2+2.5)/2=5.35\)) would be completely wrong here.

(c)
First, find the interval within \(1\) standard deviation of the mean:
\(\bar{x} \pm s = 8.2 \pm 7.2 \implies (1.0,\ 15.4)\)
Since teaching years are recorded as integers, the values that fall within this interval are \(1, 2, 3, \ldots, 15\) (since \(15 \leq 15.4 < 16\)).
From the High School B histogram, we count the teachers in the five histogram bars that contain years \(1\) through \(15\):

\((1,4):\; 79 \text{ teachers}\)
\((4,7):\; 34 \text{ teachers}\)
\((7,10):\; 28 \text{ teachers}\)
\((10,13):\; 29 \text{ teachers}\)
\((13,16):\; 19 \text{ teachers} \quad \text{(years 13, 14, 15 all} \leq 15.4\text{)}\)
\(\text{Total teachers with year} \in [1,\, 15] = 79 + 34 + 28 + 29 + 19 = 189\)
\(P(\text{within 1 SD of mean}) = \frac{189}{221}\)
\(\boxed{P \approx 0.8552}\)
The critical point here is that you must count directly from the histogram — the Empirical Rule (\(\approx 68\%\)) does not apply because High School B is strongly right-skewed, not bell-shaped.
Applying the Empirical Rule would give the wrong answer. Also note that the entire bar for \([13, 16)\) is included, because all integers in that bar (13, 14, 15) satisfy \(\leq 15.4\); only 16 would fall outside the interval, and there are no teachers listed as exactly year 16 in the \([13,16)\) bar.