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Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that a random selected student score is greater than 76

User Shutsman
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Answer:

The probability that a random selected student score is greater than 76 is
\\ P(x>76) = 0.99865.

Explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two parameters, the population mean
\\ \mu and the population standard deviation
\\ \sigma.

To determine probabilities for the normal distribution, we can use the standard normal distribution, whose parameters' values are
\\ \mu = 0 and
\\ \sigma = 1. However, we need to "transform" the raw score, in this case x = 76, to a z-score. To achieve this we use the next formula:


\\ z = (x - \mu)/(\sigma) [1]

And for the latter, we have all the required information to obtain z. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

The probability that a randomly selected student score is greater than 76

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:


\\ \mu = 85


\\ \sigma = 3


\\ x = 76

From equation [1], we have:


\\ z = (76 - 85)/(3)

Then


\\ z = (-9)/(3)


\\ z = -3

Second: Interpretation of the previous result.

In this case, the value is three (3) standard deviations below the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of
\\ z = -3, we can obtain this probability consulting the cumulative standard normal distribution, available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the cumulative standard normal distribution are for positive values of z. Fortunately, since the normal distributions are symmetrical, we can find the probability of a negative z having into account that (for this case):


\\ P(z>-3) = 1 - P(z>3) = P(z<3)

Then

Consulting a cumulative standard normal table, we have that the cumulative probability for a value below than three (3) standard deviations is:


\\ P(z<3) = 0.99865

Thus, "the probability that a random selected student score is greater than 76" for this case (that is,
\\ \mu = 85 and
\\ \sigma = 3) is
\\ P(x>76) = P(z>-3) = P(z<3) = 0.99865.

As a conclusion, more than 99.865% of the values of this distribution are above (greater than) x = 76.

We can see below a graph showing this probability.

As a complement note, we can also say that:


\\ P(z<-3) = 1 - 0.99865 = P(z>3)


\\ P(z<-3) = 0.00135 = P(z>3)

Which is the case for the probability below z = -3 [P(z<-3)], a very low probability (and a very small area at the left of the distribution).

Test scores of the student in a school are normally distributed mean 85 standard deviation-example-1
User Naeem Baghi
by
6.3k points
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