Sample space: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 6.1, 6.2, 6.3, 6.4, 6.5, 6.6.

So the mutiples are 1.3, 2.2, 2.6, 3.1, 3.5, 4.4, 5.3, 6.2, and 6.6

Greather than 6 is 1.6, 2.5, 3.4, 3.6, 4.3, 4.5 4.6, 5.2, 5.4, 5.5, 5.6, 6.1, 6.3, 6.4, and 6.5.

So the prop is 24/36 or 2/3=66%

4 0

2 years ago

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

Misha Larkins [42]

<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)

= 1 – 0.1587 [Using z table values]

= 0.8413

Now, probability of getting in between 77.6 and 88.4 = 0.8413 – 0.0228 = 0.8185

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

4 0

3 years ago