Scores on a test are normally distributed with a mean of 81.2 and a standard deviation of 3.6. What is the probability of a randomly selected student scoring between 77.6 and 88.4?

Answer 1

First, we need to find the z-scores of 77.6 and 88.4 as compared to the mean and standard deviation.

The equation is:

z = (x - mean)/(standard deviation)

Therefore,

z(77.6) = -1

z(88.4) = 2

Using the 68 - 95 - 99.7 rule, also known as the empirical rule, we can find out how many people fall within this range of -1<z<2.

Since we include all the people within one standard deviation of the mean, we can start off knowing that our answer is at least 68.

However, our range only includes the second standard deviation on the upper bound, therefore we have to cut the added probability that would bring us to 95 in half.

95 - 68 = 27; 27/2 = 13.5

We add this 13.5 to the already found 68 to get 81.5% ;

81.5% of students score between 77.6 and 88.4.

Answer 2

When working with a normal distribution, we can take note of a few things.

roughly 68% of data will be within 1 standard deviation of the mean

roughly 95% of data will be within 2 standard deviations of the mean

The standard deviation is just a representation of how cramped or spread out data is. A lower standard deviation means things are more packed, while a larger one means they are more spread out.

For this question, we have the following

mean: 81.2

standard deviation (sd): 3.6

low bound: 77.6

high bound: 88.4

Before doing any calculations we can have a rough estimate of the answer, so if we do something wrong we'll have a better chance of catching our mistake.

88.4-81.2 is 7.2. That's DOUBLE our standard deviation! That means on that side of our mean, we have 95%/2 = 47.5% of the data.

81.2-77.6 = 3.6 which is 1 standard deviation. That means on that side of our mean we have 34% of our data on the left side.

This means we should have an answer very close to 47.5+34 = 81.5%

For an answer where we have bounds like this, we want to use normalcdf(

a good resource is https://mathbits.com/MathBits/TISection/Statistics2/normaldistribution.htm

we need the lower bound, upper bound, mean, then standard deviation.

Plugging that in we get

normalcdf(77.6, 88.4, 81.2, 3.6) = .8185946784

You might say, That's not 81.5, that's 100 times smaller!

Well, you're right that 81.5 isn't .81859 but 81.5% IS almost the same as .81859

Using percentages you can move the decimal place two spots. We get the regular answer in terms of a number between 0 and 1 (inclusive) because it doesn't keep it in terms of a percentage, and no probability can be greating than 100%, or 1.

Overall, that makes the probability 81.85946784%

typical rounding puts it to the seconnd decimal place, so we would have 81.86%

For the actual formula that you have to do by hand, its

source: https://en.wikipedia.org/wiki/Normal_distribution#Cumulative_distribution_function%22

I hope you don't have to because this is only albegra 2, but that gives you an idea of what your calculator is doing behind the scenes