at a high school, the average grade for the english 10 provincial is 64, with a standard deviation of 10. If 20 students scored between 73 and 86 on the exam, how many students took the exam? Use z-score in your solution.
It must be assumed that the distribution of grades follows the shape of a normal distribution. The normal distibution is that of of a continuous random variable, however in this case the distribution is discrete. Therefore it is necessary to make the correction for continuity when calculating the z-scores. First you need to calculate the z-scores for 72.5 and 86.5. Can you do that?
ehmm... no
Have you been taught the following equation to find a z-score?\[z=\frac{X-\mu}{\sigma}\] If not why are you attempting this question?
=0.9 =2.2
i think
No, you did not use the X values that I gave you (72.5 and 86.5).
=0.95 =2.25
Now you need to look up the cumulative probabilities for bo these z-scores and subtract the smaller probability from the larger. This will give the decimal fraction of the total number of students sitting the exam represented by the 20 students.
Note that you should have calculated 0.85 for the smaller of the two z-scores.
excuse me?
what does cumulative probabilities mean?
The table at the following link shows cumulative probabilities for specified z-scores. http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
Have you recalculated the z-score for 72.5 ?
0.95= .8289 2.25= .9878
Your cumulative probability result for z = 2.25 is correct. However you are not taking notice of my advice that the z-score for 72.5 is 0.85. You need to find the cumulative probability for a z-score of 0.85 (not 0.95) Can you do that?
0.85=.8023
Correct. The decimal fraction of the total number of students sitting the exam represented by the 20 students is 0.9878 - 0.8023 = 0.1855. Therefore the total number of students sitting the exam represented by the 20 students is \[\frac{20}{0.1855}=you\ can\ calculate\]
=1077.8 so... 108 students?
oops, i meant to say 107.8
is that correct?
I would use your result of 108 students. Fractional students are not appropriate.
haha, that makes sense, thank you
You're welcome :)
can you help me with this question also? determine a 90% confidence interval for μ if σ=6, x=72, n=81
i don't know what formula to use
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