Tournament scores for 92 golfers are distributed normally. Two statistics from this tournament are given below.

Mean: 74
Standard Deviation: 2.5

What is the approximate percentage of golfers that scored higher than 79?

Same type of question as before.

Use the same link both oobleck and I gave you

To find the approximate percentage of golfers that scored higher than 79, we can use the properties of a normal distribution.

Step 1: Standardize the value.
We need to standardize the value of 79 by subtracting the mean and dividing by the standard deviation. The formula for standardization is:
Z = (X - μ) / σ
where:
Z is the standardized value,
X is the value we want to standardize (79 in this case),
μ is the mean (74 in this case), and
σ is the standard deviation (2.5 in this case).

Substituting the values, we get:
Z = (79 - 74) / 2.5
Z = 5 / 2.5
Z = 2

Step 2: Find the percentage using a standard normal distribution table.
Using a standard normal distribution table (also known as a Z-table) or a calculator, we can find the percentage associated with the standardized value of 2. In this case, we want to find the percentage of golfers that scored higher than 79, so we need to look up the area to the right of Z = 2 in the Z-table.

Consulting the Z-table, we find that the area to the right of Z = 2 is approximately 0.0228.

Step 3: Convert the percentage to a percentage format.
To convert the decimal to a percentage, we need to multiply the decimal by 100.
0.0228 * 100 = 2.28%

Therefore, approximately 2.28% of golfers scored higher than 79 in the tournament.