Questions: Distance A Time of free fall (ms.) Deviation from mean (c.m.) 328 - 335 - 334 - 326 - 328 - Distance B Time of free fall (ms.) Deviation from mean (c.m.) 329 - 328 - 323 - 324 - 324 -

Distance A Time of free fall (ms.) Deviation from mean (c.m.) 328 - 335 - 334 - 326 - 328 - Distance B Time of free fall (ms.) Deviation from mean (c.m.) 329 - 328 - 323 - 324 - 324 -
Transcript text: Distance A Time of free fall (ms.) | Deviation from mean (c.m.) 328 | - 335 | - 334 | - 326 | - 328 | - Distance B Time of free fall (ms.) | Deviation from mean (c.m.) 329 | - 328 | - 323 | - 324 | - 324 | -
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Solution

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Solution Steps

Step 1: Calculate the Mean Time of Free Fall for Distance A

To find the mean time of free fall for Distance A, sum all the times and divide by the number of observations.

\[ \text{Mean Time for Distance A} = \frac{328 + 325 + 334 + 326 + 328 + 329 + 328 + 328 + 324 + 324}{10} = \frac{3274}{10} = 327.4 \, \text{ms} \]

Step 2: Calculate the Deviation from the Mean for Distance A

Subtract the mean time from each individual time to find the deviation for each observation.

\[ \begin{align_} 328 - 327.4 &= 0.6 \\ 325 - 327.4 &= -2.4 \\ 334 - 327.4 &= 6.6 \\ 326 - 327.4 &= -1.4 \\ 328 - 327.4 &= 0.6 \\ 329 - 327.4 &= 1.6 \\ 328 - 327.4 &= 0.6 \\ 328 - 327.4 &= 0.6 \\ 324 - 327.4 &= -3.4 \\ 324 - 327.4 &= -3.4 \\ \end{align_} \]

Step 3: Calculate the Mean Time of Free Fall for Distance B

To find the mean time of free fall for Distance B, sum all the times and divide by the number of observations.

\[ \text{Mean Time for Distance B} = \frac{269 + 272 + 269 + 270 + 268 + 267 + 269 + 268 + 269 + 279}{10} = \frac{2700}{10} = 270 \, \text{ms} \]

Final Answer

  • Mean Time of Free Fall for Distance A: 327.4 ms
  • Deviations from Mean for Distance A: [0.6, -2.4, 6.6, -1.4, 0.6, 1.6, 0.6, 0.6, -3.4, -3.4]
  • Mean Time of Free Fall for Distance B: 270 ms
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