Scores on a test have a mean of 66 and Q3 of 81. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.
Q3 = 75 percentile
In a table in the back of your stat book labeled something like "areas under normal distribution," find the Z score equivalent. Put value in formula below.
Z = (x - μ)/ SD = (81-66)/SD
Solve for SD.
(81-66)/SD = 0.67(z score of 0.75)
15/SD= 0.67
15/0.67
SD=22.3880
To find the standard deviation of the scores on the test, we can use the concept of z-scores and the quartiles.
Q3 is the third quartile, which means 75% of the scores are below Q3. In a normal distribution, the z-score corresponding to the third quartile is approximately 0.6745.
So, we have:
z = 0.6745
We can use the formula for the z-score:
z = (x - μ) / σ
Where:
μ is the mean
σ is the standard deviation
We know that the mean (μ) is 66. Therefore, we can plug the values into the equation and solve for the standard deviation (σ):
0.6745 = (81 - 66) / σ
Simplifying the equation, we get:
0.6745 = 15 / σ
Cross-multiplying, we get:
0.6745σ = 15
Dividing both sides by 0.6745, we get:
σ ≈ 22.2
Therefore, the standard deviation of the scores on the test is approximately 22.2 (rounded to the nearest tenth).
To find the standard deviation, we need to use the formula for the z-score and the respective percentiles.
The Q3 (75th percentile) corresponds to a z-score of 0.674, which means that z(Q3) = 0.674.
We know that the mean (μ) is 66, and we assume the score distribution is approximately normal.
To find the standard deviation (σ), we'll use the formula for z-score: z = (X - μ) / σ, where X is the value we're looking for.
Plugging in the values, we have:
0.674 = (81 - 66) / σ
Solving for σ, we get:
0.674σ = 15
σ ≈ 15 / 0.674 ≈ 22.2
So, the standard deviation is approximately 22.2 when rounded to the nearest tenth.