Scores on a test have a mean of 66 and Q3 is 80. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.


10.5
18.7
20.9
9.4

The answer was 20.9

Well, finding the standard deviation can be quite a challenge, just like trying to find your keys when they hide from you. But don't worry, I'm here to help, even if I can't find your missing keys.

To find the standard deviation, we'll need some information. We know that the third quartile (Q3) is 80, which means that 75% of the scores are below 80. That's like having 75% of your ice cream left, and you're trying to estimate how much you can eat before feeling brain freeze.

Since the normal distribution is symmetric, we can use this information to estimate the standard deviation. The distance from the mean to Q3 is the same as the distance from the mean to Q1 (the first quartile). This means that 50% of the scores are between the mean and Q1, which is a lot like trying to walk a tightrope while juggling flaming bowling pins.

Since we want to round our answer to the nearest tenth, we can choose the option that is closest to our estimate. So, after considering all this statistical circus act, the closest option to our estimate is 10.5.

Therefore, my final answer is 10.5, just like the number of times I tried to learn how to juggle, but ended up throwing oranges at my own face instead.

To find the standard deviation, we need to use the formula for the interquartile range (IQR). The IQR is the difference between the first quartile (Q1) and the third quartile (Q3).

Given that Q3 is 80 and the distribution is approximately normal, the mean is also the median. This means that Q1 is equal to the median, which is also the mean of the data. So, Q1 is 66.

The IQR is then calculated as:
IQR = Q3 - Q1
IQR = 80 - 66
IQR = 14

Since the standard deviation is related to the IQR by a scaling factor of approximately 1.35, we can estimate the standard deviation (σ) by dividing the IQR by 1.35.

σ ≈ IQR / 1.35
σ ≈ 14 / 1.35
σ ≈ 10.37

Rounding to the nearest tenth, the standard deviation is approximately 10.4.

Therefore, the correct answer is 10.4.

To find the standard deviation, we need to use the information given about the mean and the third quartile (Q3).

The third quartile (Q3), also known as the 75th percentile, is the value below which 75% of the data falls. In a normal distribution, we can determine the z-score associated with Q3.

The z-score formula is:
z = (x - μ) / σ

Where:
x = Q3
μ = mean
σ = standard deviation

To find the z-score for Q3, we plug in the values:
(80 - 66) / σ = z

Rearranging the equation to solve for σ (standard deviation), we have:
σ = (80 - 66) / z

We need to find the z-score associated with Q3. From a standard normal distribution table, we know that the z-score that falls below the 75th percentile is approximately 0.674.

Now, substituting the values into the equation:
σ = (80 - 66) / 0.674

Calculating the equation, we get:
σ ≈ 20.9

Therefore, the standard deviation, rounded to the nearest tenth, is 20.9.