1. ## Percentage of Grades

On a standardized test with a normal distribution, the mean is 88. If the standard deviation is 4, the percentage of grades that would be expected to lie between 80 and 96 is closest to:
(1) 5 (3) 68
(2) 34 (4) 95

I'm not too clear on standard deviation questions.

2. The interval 80-96% is 2 standard deviations either side of the mean.

If you have learnt confidence intervals you would know that for a 95% confidence interval Z = 1.96, which is almost two standard deviations away from the mean. (For Z, mean is 0 and standard deviation is 1). So 95% would be the answer.

You could standardise the X value (the test scores) to the Z value and use a normal distribution table otherwise. To convert to Z, remember $\displaystyle Z = \frac{X - \mu}{\sigma}$

Finding half of the interval:

$\displaystyle P(88 < X < 96)$

$\displaystyle = P\left( \frac{88 - 88}{4} < Z < \frac{96 - 88}{4} \right)$

$\displaystyle = P(0 < Z < 2)$

Looking at a normal distribution table, the probability of this is 0.4772. Remember we just found the probability of the upper half of the interval in the question. So multiplying by 2 gives a probability very close to 0.95, i.e. 95%.

Hope you can understand.

3. ## ok........

Originally Posted by nzmathman
The interval 80-96% is 2 standard deviations either side of the mean.

If you have learnt confidence intervals you would know that for a 95% confidence interval Z = 1.96, which is almost two standard deviations away from the mean. (For Z, mean is 0 and standard deviation is 1). So 95% would be the answer.

You could standardise the X value (the test scores) to the Z value and use a normal distribution table otherwise. To convert to Z, remember $\displaystyle Z = \frac{X - \mu}{\sigma}$

Finding half of the interval:

$\displaystyle P(88 < X < 96)$

$\displaystyle = P\left( \frac{88 - 88}{4} < Z < \frac{96 - 88}{4} \right)$

$\displaystyle = P(0 < Z < 2)$

Looking at a normal distribution table, the probability of this is 0.4772. Remember we just found the probability of the upper half of the interval in the question. So multiplying by 2 gives a probability very close to 0.95, i.e. 95%.

Hope you can understand.
Interesting stuff.