A district mathematics test for all third graders had a normal distribution with a mean of 74 and a standard deviation of 11. What percentage of the third graders tested scored with +1 and -1 standard deviation of the mean?
It doesn't matter what µ and σ are.
Your Z table will show you that 68.3% lie within 1σ of µ.
To find the percentage of third graders who scored within +1 and -1 standard deviation of the mean, you can use the properties of the normal distribution.
Step 1: Determine the Z-score
The Z-score formula is given by:
Z = (X - μ) / σ
where X is the individual score, μ is the mean, and σ is the standard deviation.
For the upper limit (X = μ + 1σ):
Z = (X - μ) / σ = (μ + 1σ - μ) / σ = 1
For the lower limit (X = μ - 1σ):
Z = (X - μ) / σ = (μ - 1σ - μ) / σ = -1
Step 2: Find the corresponding percentage
Now that you have the Z-scores for the upper and lower limits, you can use a standard normal distribution table (also known as a Z-table) to find the percentage corresponding to these scores.
The Z-table provides probabilities for values up to a given Z-score. Since the normal distribution is symmetric, the probability of scoring within -1 and +1 standard deviation of the mean is the same.
Using the Z-table, you can find that the cumulative probability for a Z-score of -1 is approximately 0.1587, and for a Z-score of 1, it is approximately 0.8413.
Step 3: Calculate the percentage
To find the percentage of third graders who scored within +1 and -1 standard deviation of the mean, you subtract the cumulative probability for the lower limit from the cumulative probability for the upper limit and then multiply it by 100%.
Percentage = (0.8413 - 0.1587) * 100% = 0.6826 * 100% = 68.26%
Therefore, approximately 68.26% of the third graders tested scored within +1 and -1 standard deviation of the mean.