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maw [93]
3 years ago
6

Which variable is discrete

Mathematics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

C. The number of runners in a race

Step-by-step explanation:

Variables can be widely be classified as either discrete or continuous. A continuous variable is a variable which can take on any value within its domain. A continuous variable can take on fraction of units. Examples of continuous variables include;

time to finish a race - it can be 10.5 sec or 35.56 sec

temperature at start of race - the temperature could be 25.5 K, or 45.57 K

length of race in kilometers - the length could be 20 km, 22.45 km or 38.8 km

All the above variables can be measured in fraction of units and are thus continuous in nature.

On the other hand, a variable is said to be discrete if it can take on only integer values. An example in this case would be ;

the number of runners in a race. We can only have, 10, 45, 600, 565 and so on runners but never a fraction of a runner.

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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 rand
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<u>Answer:</u>

The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.

<u>Solution:</u>

Given, Scores on a test are normally distributed with a mean of 81.2  

And a standard deviation of 3.6.  

We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?

For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6  

Now probability of getting more than 88.4 = 1 - area of z – score of 88.4

\mathrm{Now}, \mathrm{z}-\mathrm{score}=\frac{88.4-\mathrm{mean}}{\text {standard deviation}}=\frac{88.4-81.2}{3.6}=\frac{7.2}{3.6}=2

So, probability of getting more than 88.4 = 1 – area of z- score(2)

= 1 – 0.9772 [using z table values]

= 0.0228.

Now probability of getting more than 77.6 = 1 - area of z – score of 77.6

\mathrm{Now}, \mathrm{z}-\text { score }=\frac{77.6-\text { mean }}{\text { standard deviation }}=\frac{77.6-81.2}{3.6}=\frac{-3.6}{3.6}=-1

So, probability of getting more than 77.6 = 1 – area of z- score(-1)

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Hence, the probability of a randomly selected student getting in between 77.6 and 88.4 is 0.8185.

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