Given that a population of scores is normally distriibuted with u=100 and o=8, determine the following:

a. The percentile rank of a score of 120
b. The perentage of scores that are below a score of 99
c. The percentage of scores that are between a score of 101 and 122
d. The percentage of scores that are between a score of 114 and 124
e. The score in the population aboe which 5% of the scores lie

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I think you will find it easy to answer all these questions, and if you play around some you should see how it all fits together.

To solve these questions, we will use the properties of the standard normal distribution, also known as the z-score.

a. The percentile rank of a score of 120:

Step 1: Calculate the z-score using the formula: z = (x - μ) / σ
Where x is the raw score, μ is the mean of the population, and σ is the standard deviation of the population.

z = (120 - 100) / 8
z = 2.5

Step 2: Find the percentile associated with the z-score. We use a standard normal distribution table or a calculator to find the area to the left of the z-score.

From the standard normal distribution table, we find that the area to the left of z = 2.5 is 0.9938.

Step 3: Convert the area to the percentile rank by multiplying by 100.

Percentile rank = 0.9938 * 100 = 99.38%
Thus, the percentile rank of a score of 120 is 99.38%.

b. The percentage of scores that are below a score of 99:

Step 1: Calculate the z-score using the formula: z = (x - μ) / σ

z = (99 - 100) / 8
z = -0.125

Step 2: Find the area to the left of the z-score using the standard normal distribution table.

From the standard normal distribution table, we find that the area to the left of z = -0.125 is 0.4502.

Step 3: Convert the area to the percentage by multiplying by 100.

Percentage of scores below 99 = 0.4502 * 100 = 45.02%
Thus, the percentage of scores below a score of 99 is 45.02%.

c. The percentage of scores that are between a score of 101 and 122:

Step 1: Calculate the z-scores for both scores using the formula: z = (x - μ) / σ

For 101:
z1 = (101 - 100) / 8
z1 = 0.125

For 122:
z2 = (122 - 100) / 8
z2 = 2.75

Step 2: Find the area to the left of each z-score using the standard normal distribution table.

From the standard normal distribution table, we find that the area to the left of z1 = 0.125 is 0.4515,
and the area to the left of z2 = 2.75 is 0.9974.

Step 3: Calculate the percentage between the two z-scores by subtracting the smaller area from the larger area.

Percentage of scores between 101 and 122 = (0.9974 - 0.4515) * 100 = 54.59%
Thus, the percentage of scores between a score of 101 and 122 is 54.59%.

d. The percentage of scores that are between a score of 114 and 124:

Follow the same steps as in part c:

For 114:
z1 = (114 - 100) / 8
z1 = 1.75

For 124:
z2 = (124 - 100) / 8
z2 = 3

From the standard normal distribution table, we find that the area to the left of z1 = 1.75 is 0.9599,
and the area to the left of z2 = 3 is 0.9987.

Percentage of scores between 114 and 124 = (0.9987 - 0.9599) * 100 = 3.88%
Thus, the percentage of scores between a score of 114 and 124 is 3.88%.

e. The score in the population above which 5% of the scores lie:

Step 1: Find the z-score that corresponds to the area of 0.95 (subtracting 0.05 from 1).

From the standard normal distribution table, we find the z-score corresponding to an area of 0.95 is 1.645.

Step 2: Using the formula: z = (x - μ) / σ, solve for x (the raw score).

1.645 = (x - 100) / 8

Rearranging the equation:
1.645 * 8 = x - 100
13.16 = x - 100
x = 113.16

Thus, the score in the population above which 5% of the scores lie is approximately 113.16.