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Assume the test scores is normally distributed with a mean of 80 and a standard deviation of 5. Use the 68-95-99.7 rule the following quantities a. the percentage of scores less than 80 is 50% b. the percentage of scores greater than 85 is what % solve for b

User Nathan Fraenkel
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1 Answer

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As the test scores are suppose to distribute normally, we can calculate probabilities with the formula:


P(X>85)=P(Z>(85-80)/(5))

Where Z is a normal distribution, with mean 0 and standard deviation 1

Therefore calculating the number on the right if the Z variable, we find


P(X>85)=P(Z>1)

Therefore, we have to calculate the probability that the standard normal distribution is greater than 1

For this, we use a normal table, like the next picture

This table is used in the following manner

You have to ook in the column in the left the number that yoy want to the probability be less than

In our case, we want the probability to be greater than 1

So, we have to do a little trick to get the correct expression

To get a probability for a normal distribution to be greater than a number, we apoly the formula


P(Z>1)=1-P(Z<1)

Or in a graph this can be represented as :

Now, returning to our problem we have\;


P(X>85)=P(Z>1)=1-P(Z<1)

The last probability on the right we will calculate using the table posted above

Searching the number 1 into the column on the left in the table, we find that the probability of this event is :


P(Z<1)=0.8413

Therefore, we can conclude that:


Assume the test scores is normally distributed with a mean of 80 and a standard deviation-example-1
Assume the test scores is normally distributed with a mean of 80 and a standard deviation-example-2
User Matthew Layton
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