The scores on a math test are normally distributed. if the mean is 72 and the standard deviation is six, what is the range of 98% of the scores?
You can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
Well, I must say, those math test scores sure have a sense of humor! Now, let's calculate the range of 98% of the scores.
In a normal distribution, approximately 98% of the data falls within two standard deviations of the mean. Since the standard deviation is six, we can work out the range of 98% of the scores by calculating two standard deviations below and above the mean.
Two standard deviations below the mean:
72 - (2 * 6) = 72 - 12 = 60
Two standard deviations above the mean:
72 + (2 * 6) = 72 + 12 = 84
So, the range of 98% of the scores is from 60 to 84. Just remember, if any of those scores fall outside this range, you might want to check for flying calculators or mathematically inclined clowns!
To find the range of 98% of the scores, we need to determine the z-scores that correspond to the 1% and 99% tails of the normal distribution.
First, we find the z-score for the 1% tail. Since the distribution is symmetric, the total area in the two tails (1% + 1%) is 2%. From a standard normal distribution table or using a calculator, we can find that the corresponding z-score for a 1% tail is approximately -2.33.
Next, we find the z-score for the 99% tail. Similarly, the total area in the two tails (1% + 1%) is 2%. The z-score for the 99% tail is the opposite of the z-score for the 1% tail, so it is approximately +2.33.
Now, we can calculate the actual scores corresponding to these z-scores using the formula:
score = mean + (z-score * standard deviation)
For the lower bound score:
lower bound score = 72 + (-2.33 * 6)
= 72 - 13.98
≈ 58.02
For the upper bound score:
upper bound score = 72 + (2.33 * 6)
= 72 + 13.98
≈ 85.98
Therefore, the range of 98% of the scores is approximately from 58.02 to 85.98.
To find the range of 98% of the scores, we will be using the concept of z-scores and the standard normal distribution.
1. Start by understanding the concept of the standard normal distribution. The standard normal distribution is a normal distribution with a mean of zero and a standard deviation of one. By using this distribution, we can determine the probability of a given value falling within a certain range.
2. Convert the given values to z-scores. The z-score is a measure of how many standard deviations a particular value is away from the mean.
To calculate the z-score for a specific value, use the formula: z = (x - μ) / σ
where:
- z is the z-score
- x is the given value
- μ is the mean
- σ is the standard deviation
In this case, the mean (μ) is 72 and the standard deviation (σ) is 6. We want to find the z-score for the upper boundary of the 98th percentile.
3. Use the Z-table or statistical software to find the z-score that corresponds to the 98th percentile. Since the question asks for the range of 98% of the scores, we want to find the z-score that corresponds to the upper boundary. In the standard normal distribution, this value is denoted as Z(0.98).
Using a Z-table or statistical software, we can find that the z-score corresponding to the 98th percentile is approximately 2.05.
4. Convert the z-score back to the original scores. Now that we have the z-score, we can convert it back to the original score using the formula: x = (z * σ) + μ.
Plug in the values:
x = (2.05 * 6) + 72.
Calculating this expression gives:
x ≈ 84.3.
Thus, the range of 98% of the scores is from the mean (72) to approximately 84.3.