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Statistics question?
The question is about 184 students taking an exam, mean = 72.3, standard deviation = 8.9
It asks how many students should receive a score 82-90. The answer states that 13.6%, or 25 students, should score 82-90. I don't understanding how we know that the scores between 82 and 90 constitute 13.6% of the scores. I know that 13.6% is the percentage of scores between the first and second standard deviation above the mean. But with the mean and SD given in the question, those scores should be between 81.2 and 90.1. Why do the scores between 82 and 80 also constitute 13.6%?
3 Answers
- Jacob ALv 51 decade agoFavourite answer
The scores between 82-90 shouldn't be 13.6%, the answer is incorrect.
The z score for 82 is: (82 - 72.3) / 8.9 = 1.09
This correlates to a probability of 0.8621 (Using the standard normal distribution table found in the back of most statistics textbooks)
The z score for 90 is: (90 - 72.3) / 8.9 = 1.99
This correlates to a probability of 0.9767
The probability then, equals (0.9767 - 0.8621) = 0.1146, or 11.46%
If a table is not given in the book that the question comes from, maybe they want you to estimate a lot and say that it's approx 1 std. deviation above the mean and approx. 2 std. deviations above the mean. That's the only reason I can think of for that answer.
- MerlynLv 71 decade ago
Assuming that the scores are normally distributed. I agree with you conclusion. I'm not sure why the answer provided is 13.6%.
For any normal random variable X with mean μ and standard deviation σ, X ~ Normal(μ, σ)
you can translate into standard normal units by:
Z = (X - μ) / σ
where Z ~ Normal(μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.
Let X be the average score of the test
X ~ Normal( 72.3 , 8.9 )
Find: P( 81.2 < X < 90.1 )
= P(( 81.2 - 72.3 ) / 8.9 ) < Z < ( 90.1 - 72.3 ) / 8.9 )
= P( 1 < Z < 2 )
= P( Z < 2 ) - P( Z < 1 )
= 0.9772499 - 0.8413447
= 0.1359051
X ~ Normal( 72.3 , 8.9 )
Find: P( 82 < X < 90 )
= P(( 82 - 72.3 ) / 8.9 ) < Z < ( 90 - 72.3 ) / 8.9 )
= P( 1.089888 < Z < 1.988764 )
= P( Z < 1.988764 ) - P( Z < 1.089888 )
= 0.9766364 - 0.8621187
= 0.1145177
- Stephen YLv 61 decade ago
essentially because the scores must be whole numbers.
I grant I would be happier between 81 and 90 but so be it.
