For a normal distribution with a mean of m = 80 and a standard deviation of s = 20, find the proportion of the population corresponding to each of the following scores.

a. Scores greater than 85
b. Scores less than 100
c. Scores between 70 and 90

I believe this is the correct formula to start with but am unclear on how to finish it. Any help would be appreciated.

a. z = X-m= 85–80 = 5 = 0.25
s 20 20

b. z = X-m= 100–80 = 20 = 1.00
s 20 20

c. z = X-m= 70–80 = -10 = -0.25
s 20 20
z = X-m= 90–80 = 10 = 0.50
s 20 20

What does "s 20 20" mean?

Z = (x-mean)/SD

Type equation completely to avoid mistakes.

c. -10/20 = ?

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores you have found.

I hope this helps.

a. For scores greater than 85, you need to find the proportion of the population to the right of that score. Since you've already calculated the z-score for 85 as 0.25, you can look up the corresponding area under the normal curve in a standard normal distribution table. The table will give you the proportion of the population to the left of that z-score. To get the proportion to the right, you subtract this value from 1. So, the proportion of scores greater than 85 is 1 - 0.5987 = 0.4013.

b. For scores less than 100, you've already calculated the z-score for 100 as 1.00. Again, you can look up the corresponding area under the normal curve in a standard normal distribution table, which will give you the proportion of the population to the left of that z-score. So, the proportion of scores less than 100 is 0.8413.

c. For scores between 70 and 90, you need to find the proportion of the population between those two scores. You've already calculated the z-scores for 70 (-0.25) and 90 (0.50). Again, you can use the standard normal distribution table to find the proportion of the population to the left of each z-score, and subtract these values to find the proportion between the two scores. So, the proportion of scores between 70 and 90 is 0.6915 - 0.4013 = 0.2902.

To find the proportion of the population corresponding to each of the given scores, we need to use the standard normal distribution table.

a. To find the proportion of scores greater than 85, we need to find the area under the curve to the right of 85. From the table, the z-score corresponding to 85 is 0.25. Using the table, we find that the proportion of scores greater than 85 is 0.5987 or 59.87%.

b. To find the proportion of scores less than 100, we need to find the area under the curve to the left of 100. The z-score corresponding to 100 is 1.00. Using the table, we find that the proportion of scores less than 100 is 0.8413 or 84.13%.

c. To find the proportion of scores between 70 and 90, we need to find the area under the curve between the z-scores corresponding to 70 and 90. The z-score corresponding to 70 is -0.25, and the z-score corresponding to 90 is 0.50. Using the table, we find the proportion of scores below 70 is 0.4013 or 40.13%, and the proportion of scores below 90 is 0.6915 or 69.15%. To find the proportion of scores between 70 and 90, we subtract the proportion below 70 from the proportion below 90: 0.6915 - 0.4013 = 0.2902 or 29.02%.

Therefore:

a. The proportion of the population with scores greater than 85 is 59.87%.
b. The proportion of the population with scores less than 100 is 84.13%.
c. The proportion of the population with scores between 70 and 90 is 29.02%.

To find the proportion of the population corresponding to each of the given scores, we need to use the standard normal distribution table (also known as the Z-table) or a statistical software.

Let's go through each part one by one:

a. Scores greater than 85:
From the given information, we have calculated z-score using the formula: z = (X - m) / s, where X is the score, m is the mean, and s is the standard deviation. For this case, z = (85 - 80) / 20 = 0.25.

To find the proportion of scores greater than 85, we need to find the area under the normal curve to the right of the z-score of 0.25. The easiest way to do this is by using a standard normal distribution table.

From the standard normal distribution table, the area to the right of z = 0.25 is 0.4013. This represents approximately 40.13% of the population.

b. Scores less than 100:
For this case, the z-score is calculated as z = (100 - 80) / 20 = 1.00.

To find the proportion of scores less than 100, we need to find the area under the normal curve to the left of the z-score of 1.00.

From the standard normal distribution table, the area to the left of z = 1.00 is 0.8413. This represents approximately 84.13% of the population.

c. Scores between 70 and 90:
We have two values to consider: 70 and 90.

For 70, the z-score is calculated as z = (70 - 80) / 20 = -0.50.
For 90, the z-score is calculated as z = (90 - 80) / 20 = 0.50.

To find the proportion of scores between 70 and 90, we need to find the area under the normal curve between these two z-scores.

From the standard normal distribution table, the area to the left of z = -0.50 is 0.3085. (This represents the proportion of scores less than 70).
The area to the left of z = 0.50 is 0.6915. (This represents the proportion of scores less than 90).

To find the proportion of scores between 70 and 90, we subtract the smaller area from the larger area: 0.6915 - 0.3085 = 0.3830. This represents approximately 38.30% of the population.

So, the proportions of the population for the given scores are:
a. Scores greater than 85: 40.13%
b. Scores less than 100: 84.13%
c. Scores between 70 and 90: 38.30%