Suppose the final exam scores in a large class is symmetric mound shaped with a mean of 70 points and standard deviation of 10 points. The percentage of students in the class who receive more than 60 points is approximately

60% is 70%-10% (1 standard deviation).

Look up a normal distribution table (or
http://www.mathsisfun.com/data/standard-normal-distribution-table.html
)
to find that at Z=1 (1 standard deviation) below mean is (0.5)-0.3413=0.1587
So 15.87% of the class has (70-10)=60% or less. So how many are above 60?

Use z-scores.

Formula:
z = (x - mean)/sd

x = 60
mean = 70
sd = 10

Substitute into the formula and find z. Use a z-distribution table to find the probability. Convert to a percent.

I hope this will help get you started.

Hmm, let me calculate that for you. Ah, here we go! Since the scores are normally distributed, we can use the empirical rule to estimate the percentage of students who score above 60 points.

According to the empirical rule, approximately 68% of the scores fall within one standard deviation of the mean. So, that means around 68% – or should I say, "a whole lot" – of students should score between 60 and 80 points.

But we want to estimate the percentage of students who score above 60 points, not between 60 and 80. So, we need to find out how many scores fall within one standard deviation below the mean (60 points) and subtract it from 68%.

To do that, we calculate the z-score for 60 points using the formula: z = (x - mean) / standard deviation.

In this case, z = (60 - 70) / 10 = -1. So, we need to find the percentage of scores below -1 z-score.

Consulting my magical statistical guidebook, I see that about 15.87% of scores fall below -1 z-score.

Now, we simply subtract this percentage from 68%. So, 68% - 15.87% = 52.13% (approximately).

Therefore, the percentage of students in the class who receive more than 60 points is approximately 52.13%.

To find the percentage of students in the class who receive more than 60 points, we can use the z-score formula and the standard normal distribution table.

The z-score formula is given by:

z = (x - μ) / σ

Where:
z = z-score
x = raw score
μ = mean
σ = standard deviation

Given that the mean (μ) is 70 points and the standard deviation (σ) is 10 points, let's calculate the z-score for a raw score of 60 points:

z = (60 - 70) / 10
z = -1

Now, we can use the standard normal distribution table to find the percentage of students who score more than 60 points. The table provides the area under the curve to the left of a given z-score.

Looking up the z-score of -1 in the table, we find that the area to the left of -1 is approximately 0.1587.

To find the percentage of students who score more than 60 points, we subtract this area from 1:

Percentage = 1 - 0.1587
Percentage = 0.8413

Therefore, approximately 84.13% of the students in the class will receive more than 60 points on the final exam.

To find the percentage of students in the class who receive more than 60 points, we can use the standard normal distribution.

First, we need to convert the given values (mean and standard deviation) to the standard units, or z-scores. The formula to compute the z-score is:

z = (x - μ) / σ

where x is the value, μ is the mean, and σ is the standard deviation.

For the given problem, we have x = 60, μ = 70, and σ = 10.

So, the z-score is:

z = (60 - 70) / 10
= -1

Now, we need to find the area under the standard normal curve to the right of z = -1. This represents the percentage of students who scored more than 60 points.

Using a standard normal distribution table or a calculator, we find that approximately 0.8413 of the area lies to the left of z = -1. Therefore, the area to the right (the percentage of students who scored more than 60 points) is approximately:

1 - 0.8413 = 0.1587

To convert this to a percentage, we multiply by 100:

0.1587 * 100 = 15.87%

Therefore, approximately 15.87% of the students in the class will receive more than 60 points on the final exam.