1. if a 1,000 students take a test that has a mean of 40 mins. a standard deviation of 8 mins. and is normally distributed how many would you expect to finish in less than 40 mins     ?

2. The time required to finish a test is normally distributed with a mean 80 minutes     and a standars deviation of 15 minutes. what is the probability that a student chosen at random will finish the test between 50 and 95 mins?

3. the time required to finish a test is normally distributed with a mean of 40 mins. and a standard deviation of 8 mins. what is the probability that a student chosen at random will finish the test in less than 32 mins?

4. the average score on a standardized test is 750 points with a standard deviation of 50 points. what is the probabilty that the student scores between 450 and 600 on the standardized test ?

5. the average score on a standardized test is 750 points with a standard deviation of 50 points what is the probability that the student scores more than 700 points on the standardized test ?

6. Suppose you have a mean standardized test score 1200 points with a standard deviation of 200 points. this data is normally distributed what is the z-score of 900points?

7. suppose you have a normally distributed set of data pertaining to standardized test. the mean score is 1000 and the standard deviation is 200. what is the z-score of 900 points score.

8. suppose you have a mean standardized score of 1500 points with a standard deviation of 150 points. this data is normally distributed. what is the Z-score of 1600 points.

9. suppose you have a mean standardized score of 1500 points with a standard deviation of 150 points. this data is normally distributed. what is the z-score of1225 points?

1)

Number of students,n=1000

Mean time ,mue=40 min

standard deviation,sigma=8

From normal distributed table,

At N(40,8)

f(z)=0.5

Expected numbers=nf(z)

=0.5*1000

=500

Mean,mue=80 min

Standard deviation,sigma=15 min

x1=50

x2=95

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probability of a student chosen at random will finish the test between 50 and 95 mins

P((x1-mue)/sigma<=z<=(x2-mue)/sigma)=P((50-80)/15<=z<=(95-80)/15)

=P(-2<=z<=1)

=P(z<=1)-P(z<=-1)

=0.85314-0.2018       (Since from normal distributive table)

=0.65134