SOLUTION: the mean test score of 50 students in a statistics class is 80 with a standard deviation of 7. if the test scores known to be normally distributed, calculate the following. 1. wha

Question 1162116: the mean test score of 50 students in a statistics class is 80 with a standard deviation of 7. if the test scores known to be normally distributed, calculate the following.
1. what proportion of the students who scored between 80 and 90?
2. what percentage of the students who scored from 60 to 80?
3. what is the probability of the students who scored between 70 and 85?
4. how many students have scored between 85 and 96?
5. how many students obtained a score range from 68 to 75?
6. howmany students who scored greater than 93?
7. How many of the students scored less than 65?
8.find the percentage of the students who scored less than 90?
9.if 30% of the students fail, what is the passing score?
10. Find the range of score that covers the central 78.88% of the students?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd
=(80-80)/7=0
=(90-80)/7=1.43
want z between 0 and 1.43, which is 0.4236
OR, and you can do this with all, go to 2nd VARS 2 Normalcdf (80,90,80,7). put in the two limits followed by the mean and the sd. ENTER.
so for 4. put in for normalcdf (85,96,80,7)ENTER 0.2264 probability.
for 6, put in a large number like 1000. normal cdf (93,1000,80,7)
for 9. use 2ndENTER 2 invnorm (0.30,80,7) ENTER. Get 76.3
z(0.30)=-0.5244
z=(x-mean)/sd
-0.524*7=x-mean
x=80-3.67=76.3
there is 21.12% remaining, (100-78.88) and half of that is 10.56%
inv norm (0.1056,50,7)=41.25
that is 8.75 below the mean. The normal function is symmetrical, so the other side will be 8.75 above the mean, and that would be 58.75.
Check with normal cdf (41.25, 58.75, 50, 7)=0.7887, which is off by rounding error. Use 3 decimal places.

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