Find the standard deviation for the data set. 35,36,36,38,41,42,45,48?

Answer 1

See a solution process below:

Step 1) Work out the Mean (the simple average of the numbers)

#m_n = (35 + 36 + 36 + 38 + 41 + 42 + 45 + 48)/8#
#m_n = 321/8#
#m_n = 40.125#

Step 2) Then for each number: subtract the Mean and square the result

#(35 - 40.125)^2 = 26.265625#
#(36 - 40.125)^2 = 17.015625#
#(36 - 40.125)^2 = 17.015625#
#(38 - 40.125)^2 = 4.515625#
#(41 - 40.125)^2 = 0.765625#
#(42 - 40.125)^2 = 3.515625#
#(45 - 40.125)^2 = 23.765625#
#(48 - 40.125)^2 = 62.015625#

Step 3) Then work out the mean of those squared differences.

#m_s = (26.265625 + 17.015625 + 17.015625 + 4.515625 + 0.765625 + 3.515625 + 23.765625 + 62.015625)/8#
#m_s = 154.878/8#
#m_s = 19.35975#

Step 4) Take the square root of that and we are done!

#s = sqrt(19.35975)#
#s = 4.399971590817377#
#s = 4.4# rounded to the nearest tenth
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Answer 2

To find the standard deviation for the given data set, you can follow these steps:

  1. Find the mean (average) of the data set.
  2. Calculate the difference between each data point and the mean.
  3. Square each of these differences.
  4. Find the mean of the squared differences.
  5. Take the square root of the mean squared differences to find the standard deviation.

Using the provided data set: 35, 36, 36, 38, 41, 42, 45, 48

  1. Mean ( \bar{x} = \frac{35 + 36 + 36 + 38 + 41 + 42 + 45 + 48}{8} = \frac{321}{8} = 40.125 )

  2. Differences from the mean: (35 - 40.125 = -5.125), (36 - 40.125 = -4.125), (36 - 40.125 = -4.125), (38 - 40.125 = -2.125), (41 - 40.125 = 0.875), (42 - 40.125 = 1.875), (45 - 40.125 = 4.875), (48 - 40.125 = 7.875).

  3. Square of each difference: ( (-5.125)^2 = 26.265625), ( (-4.125)^2 = 17.015625), ( (-4.125)^2 = 17.015625), ( (-2.125)^2 = 4.515625), ( (0.875)^2 = 0.765625), ( (1.875)^2 = 3.515625), ( (4.875)^2 = 23.765625), ( (7.875)^2 = 62.015625).

  4. Mean of the squared differences: ( \frac{26.265625 + 17.015625 + 17.015625 + 4.515625 + 0.765625 + 3.515625 + 23.765625 + 62.015625}{8} = \frac{155.875}{8} = 19.484375 ).

  5. Standard deviation: ( \sqrt{19.484375} \approx 4.41 ).

Therefore, the standard deviation for the given data set is approximately 4.41.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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