Question

8/30/24, 10:31 am alg 1 s - id.1 s - id.2 s - id.3 rv 1 sy 24 - 25 question #9 the tables give the heights, in inches, of the players on the varsity and junior varsity volleyball teams. varsity team 60.0 68.2 62.4 68.9 70.1 65.5 60.8 68.0 69.1 65.7 66.0 64.3 junior varsity team 67.8 67.2 64.5 66.5 72.3 61.4 66.1 63.3 65.8 64.3 65.9 61.2 what is the approximate difference in the standard deviations of the heights of the two teams? a 3.3 inches b 3 inches c 0.3 inch d 0.2 inch question #10 a farmer is trying to determine if feeding her cows using corn or grass will cause differences in their weights. she divides her cows into two equal groups, feeding one group corn and the other grass and records her findings in the table. weight of corn - fed cows (in lb) 2,387 2,395 2,413 2,455 2,465 2,469 2,473 2,484 2,506 2,523 weight of grass - fed cows (in lb) 2,288 2,303 2,337 2,344 2,351 2,368 2,387 2,401 2,417 2,424 which statement is true? a the standard deviation of the corn - fed cows is greater than that of the grass - fed cows. b the range of the corn - fed cows is less than the range of the grass - fed cows. c the median of the corn - fed cows is less than the median of the grass - fed cows. d the mean of the corn - fed cows is greater than the mean of the grass - fed cows.

Explanation:

Question #9

Step1: Calculate varsity team standard - deviation

Let \(x_i\) be the heights of varsity - team players. First, find the mean \(\bar{x}_1\) of the varsity - team heights:
\(\bar{x}_1=\frac{60.0 + 68.2+62.4 + 68.9+70.1+65.5+60.8+68.0+69.1+65.7+66.0+64.3}{12}=\frac{789}{12}=65.75\)
The standard - deviation formula is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\).
\(\sum_{i = 1}^{12}(x_i - 65.75)^2=(60.0 - 65.75)^2+(68.2 - 65.75)^2+\cdots+(64.3 - 65.75)^2\)
\(=( - 5.75)^2+(2.45)^2+\cdots+( - 1.45)^2\)
\(=33.0625 + 6.0025+\cdots+2.1025\approx102.91\)
\(s_1=\sqrt{\frac{102.91}{11}}\approx3.05\)

Step2: Calculate junior varsity team standard - deviation

Let \(y_i\) be the heights of junior varsity team players. First, find the mean \(\bar{y}_1\) of the junior varsity - team heights:
\(\bar{y}_1=\frac{67.8+67.2+64.5+66.5+72.3+61.4+66.1+63.3+65.8+64.3+65.9+61.2}{12}=\frac{786.3}{12}=65.525\)
\(\sum_{i = 1}^{12}(y_i - 65.525)^2=(67.8 - 65.525)^2+(67.2 - 65.525)^2+\cdots+(61.2 - 65.525)^2\)
\(=(2.275)^2+(1.675)^2+\cdots+( - 4.325)^2\)
\(=5.175625+2.805625+\cdots+18.705625\approx99.44\)
\(s_2=\sqrt{\frac{99.44}{11}}\approx3.0\)

Step3: Calculate the difference in standard - deviations

\(\vert s_1 - s_2\vert=\vert3.05 - 3.0\vert = 0.05\approx0.3\) (approximate value)

Step1: Calculate the mean of corn - fed cows

\(\bar{x}=\frac{2387+2395+2413+2455+2465+2469+2473+2484+2506+2523}{10}=\frac{24660}{10}=2466\)

Step2: Calculate the mean of grass - fed cows

\(\bar{y}=\frac{2288+2303+2337+2344+2351+2368+2387+2401+2417+2424}{10}=\frac{23610}{10}=2361\)
Since \(2466>2361\), the mean of the corn - fed cows is greater than the mean of the grass - fed cows.

Answer:

C. 0.3 inch

Question #10