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dangina [55]
3 years ago
14

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

Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

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

Step-by-step explanation:

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

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

\\ z = \frac{x - \mu}{\sigma} [1]

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

<h3>The probability that a randomly selected student score is greater than 76</h3>

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 = \frac{76 - 85}{3}

Then

\\ z = \frac{-9}{3}

\\ z = -3

Second: Interpretation of the previous result.

In this case, the value is <em>three</em> (3) <em>standard deviations</em> <em>below</em> 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 <em>the cumulative standard normal distribution, </em>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 <em>cumulative standard normal distribution</em> are for positive values of z. Fortunately, since the normal distributions are <em>symmetrical</em>, 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

Then

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

\\ P(z

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.

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

<em>We can see below a graph showing this probability.</em>

As a complement note, we can also say that:

\\ P(z3)

\\ P(z3)

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).

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