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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.
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
Finding half of the interval:
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.Quote:
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
Finding half of the interval:
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.