Given a population of scores is normally distributed with mu of 110 and a standard deviation of 8 the percentage of scores that are below a score of 99?
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 related to the Z score.
To find the percentage of scores that are below a score of 99, we need to calculate the z-score first.
The z-score formula is:
z = (x - μ) / σ
Where:
x = score value
μ = mean (average) of the population
σ = standard deviation of the population
Now let's plug in the values:
x = 99
μ = 110
σ = 8
z = (99 - 110) / 8
z = -11 / 8
z ≈ -1.375
Once we have the z-score, we can use a standard normal distribution table or calculator to find the corresponding percentile. The percentile represents the percentage of scores that are below a given z-score.
Looking up the z-score of -1.375 in a standard normal distribution table, we find the percentile to be approximately 0.0838.
To find the percentage, we multiply the percentile by 100:
Percentage = 0.0838 * 100
Percentage ≈ 8.38%
Therefore, approximately 8.38% of scores are below a score of 99.
To find the percentage of scores that are below a score of 99 in a normal distribution, you can use the standard normal distribution table or a statistical calculator.
Here are the steps to calculate it using the standard normal distribution table:
1. Convert the given score of 99 to a z-score. The z-score measures the number of standard deviations a particular score is away from the mean. The formula to calculate the z-score is:
z = (x - mu) / sigma
where x is the score, mu is the mean, and sigma is the standard deviation. In this case:
z = (99 - 110) / 8
= -11 / 8
= -1.375
2. Look up the z-score (-1.375) in the standard normal distribution table. The table provides the proportions (or percentages) of scores below a particular z-score. In this case, you would look up the value closest to -1.375 and find its corresponding percentage.
The standard normal distribution table typically gives the area/proportion to the left of the z-score. So, if you find the closest z-score, you will get the proportion of scores below that value.
Let's assume that the closest z-score in the table is -1.38. The corresponding proportion/percentage for this z-score is 0.0838.
3. The proportion obtained from the table represents the area to the left of the z-score. To find the percentage of scores below the given score of 99, you can convert this proportion to a percentage by multiplying it by 100.
percentage = proportion * 100
= 0.0838 * 100
= 8.38%
So, approximately 8.38% of the scores are below a score of 99 in this normal distribution with a mean of 110 and a standard deviation of 8.
Note: If you have access to a statistical calculator or software, you can also directly input the score, mean, and standard deviation to get the desired result.