The chart above represents the number of students at five schools, eac

Solution 1.

School 1:
\(\overline{x_1} = \frac{148+151+153+148}{4}=\frac{600}{4}=150\)
\(\sigma^2_1=(148-150)^2 + (151-150)^2 + (153-150)^2 + (148-150)^2=2^2+1+3^2+2^2\)

School 2:
\(\overline{x_2} = \frac{156+99+148+157}{4}=\frac{560}{4}=140\)
\(\sigma^2_2=(156-140)^2 + (99-140)^2 + (148-140)^2 + (157-140)^2=16^2+41^2+8^2+17^2 > \sigma^2_1\)

School 3:
\(\overline{x_3} = \frac{104+120+196+140}{4}=\frac{560}{4}=140\)
\(\sigma^2_3=(104-140)^2 + (120-140)^2 + (196-140)^2 + (140-140)^2=36^2+20^2+56^2+0 > \sigma^2_1\)

School 4:
\(\overline{x_4} = \frac{95+156+204+35}{4}=\frac{490}{4}=122.5\)
\(\sigma^2_4=(95-122.5)^2 + (156-122.5)^2 + (204-122.5)^2 + (35-122.5)^2 > \sigma^2_1\)

School 5:
\(\overline{x_5} = \frac{217+74+159+220}{4}=\frac{670}{4}=167.5\)
\(\sigma^2_5=(217-167.5)^2 + (74-167.5)^2 + (159-167.5)^2 + (220-167.5)^2> \sigma^2_1\)

Hence the least standard deviation is from school 1. The answer is A.

Solution 2.

School 1. \(\{148;148;151;153\}\)
School 2. \(\{99;148;156;157\}\)
School 3. \(\{104;120;140;196\}\)
School 4. \(\{35;95;156;204\}\)
School 5. \(\{74;159;217;220\}\)

It's clear that the difference between any two elements in school 1 is the least. Hence, the standard deviation in school 1 will remain the least. The answer is A.