The mean score on the first calculus exam was 55%. The standard deviation was 7%. Find the smallest interval about the mean that must contain at least 83% of the scores.
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Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.415) related to the Z scores. Insert Z values into the above equation.
did u figure this out yet....? ahhhh
To find the smallest interval about the mean that must contain at least 83% of the scores, we need to use the empirical rule or the 68-95-99.7 rule.
According to the empirical rule, for a normal distribution:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
In this case, we know that the mean score was 55% and the standard deviation was 7%. We want to find the interval that contains at least 83% of the scores.
To find this interval, we need to find the number of standard deviations needed to capture at least 83% of the data. We can use the formula:
Z = (X - μ) / σ
Where:
Z is the number of standard deviations from the mean
X is the score we want to find the interval for
μ is the mean score
σ is the standard deviation
To find the Z value corresponding to 83% of the data:
Z = invNorm(0.83)
Using a calculator or statistical software, we can find that Z ≈ 0.983.
Now, we can set up the following equation:
0.983 = (X - 55) / 7
Solving for X, we get:
X - 55 = 0.983 * 7
X - 55 ≈ 6.881
X ≈ 55 + 6.881
X ≈ 61.881
Therefore, the smallest interval about the mean that must contain at least 83% of the scores is approximately (55 - 61.881) to (55 + 61.881), or (approximately) (-6.881 to 116.881).