The test result for maths are normally destributed . The mean is 46 % and the standerd deviation is 4. What percentage of student obtains a mark :
a) less than 49
B) within 1.8 standard deviations from the mean
A. The math is right, interpretation is not. If your mean is 46, and you are being asked for scores LESS THAN a statistic that is higher than the mean (49), then your answer should be at least 50%. Do you see why? If the test scores are normal (and they are), then 50% of the scores will be less than the mean, and 50% will be higher than the mean. You are being asked for scores less than a score higher than the mean, which means you are being asked for the area under the normal curve to the left of your score (the 49). What kind of z-table are you using. If you are using one that just has one half of the table (meaning it goes from 0 to 3.5), then you have to be careful on how you use that, as it is not a CUMMULATIVE z-score table. In your case, you need to find 0.75 and the percentage associated with it (which is indeed 0.2734) and then ADD .5 to it.
B. For part B we are being asked to find the percentage of scores that are within 1.8 standard deviations from the mean. Note that they did not tell us above or below, but simply within. That means we have to find two statistics. Remember what a z-score is; it is the number of standard deviations a statistic is away from the mean, a mean that is part of a normal population. So if we are being asked for test statistics 1.8 within the mean, that means we need to find the statistics that will give us a z-score of plus or minus 1.8:
You can take it from here.
For the question a i understood why add the .5 because of the 50 50 around the mean
for the question b i need the stats X so rearaging the furmula around the x the question gives the z score of 1,8
So for the Z= 1.8 the Stats will be 53.2
and the Z-1.8 the stats will be 38.8
thanks for the help .. very much appreciated ..
For problem b, you are already given the z score! Some normal distribution tables just give the probability from 0 to z. With one of those, look up the probability for z between 0 and 1.8 and double it. That uses the fact that the normal curve is symmetric about z= 0.
Others give the probability from negative infinity to z, but only for z positive. With one of those, look up the probability for z= 1.8, double it and subtract 1 from that. That's because, if the probability from negative infinity to z is P, then the probability from z to positive infinity is 1- P. Because the normall distribution is symmetric, that is also the probability from negative infinity to -z and that you want to subtract off: P-(1-P)= 2P- 1.