When Mrs. Myles gave a test, the scores were normally distributed with a mean of 72 and a standard deviation of 8. This means that 95% of her students scored between which two scores?
A)
40 and 100
B)
48 and 96
C)
56 and 88
D)
64 and 80
Well, I hope none of Mrs. Myles' students were clowns because according to the normal distribution, 95% of the scores fall within two standard deviations from the mean. So, let me use my funny calculations to give you the answer.
The mean is 72 and the standard deviation is 8. Two standard deviations above and below the mean would be 16 points in total. So, the range would go from 72 - 16 to 72 + 16.
That means the range is from 56 to 88. So, the students' scores would be between 56 and 88. That narrows it down to answer choice C!
To find the range of scores between which 95% of Mrs. Myles' students scored, we need to calculate the interval that falls within 2 standard deviations from the mean.
The Z-score formula is used to transform the normal distribution into a standard normal distribution (mean = 0, standard deviation = 1).
The formula for the Z-score is given as:
Z = (X - μ) / σ
Where:
Z = Z-score
X = Individual score
μ = Mean
σ = Standard deviation
To find the interval, we need to find the Z-scores that correspond to the lower and upper limits of the range.
For the lower limit (Z1), we use the formula:
Z1 = (X1 - μ) / σ
Since we want to find the score that corresponds to the lower 2.5% (0.025) of the distribution (as 95% falls within 2 standard deviations from the mean), we find the value of Z at this percentile using a standard normal distribution table.
From the table, we find that Z1 = -1.96 (approximately).
Using the Z-score formula:
-1.96 = (X1 - 72) / 8
Rearranging the formula to solve for X1 (the lower score):
X1 = -1.96 * 8 + 72 = 56.32
Similarly, for the upper limit (Z2), we find the Z-score that corresponds to the upper 2.5% (0.975) of the distribution. Again, from the standard normal distribution table, we find Z2 = 1.96 (approximately).
Using the Z-score formula:
1.96 = (X2 - 72) / 8
Rearranging the formula to solve for X2 (the upper score):
X2 = 1.96 * 8 + 72 = 87.68
Thus, 95% of Mrs. Myles' students scored between approximately 56.32 and 87.68.
The answer is C) 56 and 88.
To find the two scores between which 95% of the students scored, we can use the empirical rule for a normal distribution.
According to the empirical rule, for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations.
In this case, the mean is 72 and the standard deviation is 8.
To find the scores between which 95% of the students scored, we need to consider two standard deviations on either side of the mean.
Two standard deviations below the mean would be: 72 - (2 * 8) = 72 - 16 = 56
Two standard deviations above the mean would be: 72 + (2 * 8) = 72 + 16 = 88
Therefore, 95% of the students scored between 56 and 88.
Hence, the correct answer is C) 56 and 88.