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SajjadAhmad
The arithmetic mean of a data set is 46 and the standard deviation of the set is 4. Which of the following contains the interval two standard deviations from the mean of the set?

(A) 38 to 46
(B) 38 to 54
(C) 42 to 50
(D) 44 to 48
(E) 46 to 50
\(Range\) \(=\) \(Mean\) + \(2*SD\)

Or, \(Range\) \(=\) \(46\) + \(2*4\)

Or, \(Range\) \(=\) \(46\) + \(8\)

Or, \(Range\) \(=\) \(38 \ to \ 54\)

Hence, the correct answer must be (B) \(38 to 54\)
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SajjadAhmad
The arithmetic mean of a data set is 46 and the standard deviation of the set is 4. Which of the following contains the interval two standard deviations from the mean of the set?

(A) 38 to 46
(B) 38 to 54
(C) 42 to 50
(D) 44 to 48
(E) 46 to 50

Two standard deviations above the mean is 46 + 2(4) = 46 + 8 = 54.

Two standard deviations below the mean is 46 - 2(4) = 46 - 8 = 38.

Answer: B
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SajjadAhmad
The arithmetic mean of a data set is 46 and the standard deviation of the set is 4. Which of the following contains the interval two standard deviations from the mean of the set?

(A) 38 to 46
(B) 38 to 54
(C) 42 to 50
(D) 44 to 48
(E) 46 to 50

-----------ASIDE-----------------
A little extra background on standard deviations above and below the mean

If, for example, a set has a standard deviation of 4, then:
1 standard deviation = 4
2 standard deviations = 8
3 standard deviations = 12
1.5 standard deviations = 6
0.25 standard deviations = 1
etc


So, if the mean of a set is 9, and the standard deviation is 4, then:
2 standard deviations ABOVE the mean = 17 [since 9 + 2(4) = 17]
1.5 standard deviations BELOW the mean = 3 [since 9 - 1.5(4) = 3]
3 standard deviations ABOVE the mean = 21 [since 9 + 3(4) = 21]
etc.
----------------------------------

The arithmetic mean of a data set is 46 and the standard deviation of the set is 4. Which of the following contains the interval two standard deviations from the mean of the set?

2 standard deviations ABOVE the mean = 46 + 2(4) = 54
2 standard deviations BELOW the mean = 46 - 2(4) = 38

The range = 38 to 54

Answer: B

Cheers,
Brent
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