Normal Distribution! HELP AGAIN!

ckpridham

New member
Joined
Apr 22, 2013
Messages
2
I MISSED THIS PART OF THE LECTURE! I had to go to the hospital, my lack of knowledge in this subject is not due to laziness. I have everything else down but normal distribution and plan on going to the success center at my college as soon at it opens at 3 for math. Until then if I could get a little insight I would REALLY appreciate it! Thank you!!!


4.
The mean score on a standardized aptitude test was 100 with a standard deviation of 10. Assume the scores are normally distributed. The percentage of the people who took this test and had a score between 102 and 112 is ________%. You only need to submit a numerical value - do not include the % sign with your answer. Round your answer to 1 decimal place.



5.
The mean height for a group of children is 48.1 inches with a standard deviation of 3.2 inches. Assume these heights are normally distrbitued. If one child from this group is selected at random, then the probability their height is between 52 inches and 56 inches is ________.
Round your answer to 3 decimal places.











One method of "grading on a curve" uses the normal distribution curve with the scheme shown in the table below, where z is the number of standard deviations a score is from the mean.

The mean on an exam in a large section class of 120 students was 73.9 with a standard deviation of 12.8. Approximately how many students will get a grade of B or higher? [If a decimal occurs, round your answer up to the next whole person.]

Score
Grade
z > 1.5​
A​
0.5 < z < 1.5​
B​
−0.5 < z < 0.5​
C​
−1.5 < z < −0.5​
D​
z < −1.5​
F​

approximate number of students:




7.
Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Answer the following based on this information rounding your answer to 3 decimal places.
Find the probability a randomly selected adult has an IQ scores that is greater than 140.



 
I MISSED THIS PART OF THE LECTURE! I had to go to the hospital, my lack of knowledge in this subject is not due to laziness. I have everything else down but normal distribution and plan on going to the success center at my college as soon at it opens at 3 for math. Until then if I could get a little insight I would REALLY appreciate it! Thank you!!!


4.
The mean score on a standardized aptitude test was 100 with a standard deviation of 10. Assume the scores are normally distributed. The percentage of the people who took this test and had a score between 102 and 112 is ________%. You only need to submit a numerical value - do not include the % sign with your answer. Round your answer to 1 decimal place.




5.
The mean height for a group of children is 48.1 inches with a standard deviation of 3.2 inches. Assume these heights are normally distrbitued. If one child from this group is selected at random, then the probability their height is between 52 inches and 56 inches is ________.
Round your answer to 3 decimal places.










One method of "grading on a curve" uses the normal distribution curve with the scheme shown in the table below, where z is the number of standard deviations a score is from the mean.

The mean on an exam in a large section class of 120 students was 73.9 with a standard deviation of 12.8. Approximately how many students will get a grade of B or higher? [If a decimal occurs, round your answer up to the next whole person.]


Score

Grade

z > 1.5​

A​

0.5 < z < 1.5​

B​

−0.5 < z < 0.5​

C​

−1.5 < z < −0.5​

D​

z < −1.5​

F​

approximate number of students:




7.
Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Answer the following based on this information rounding your answer to 3 decimal places.
Find the probability a randomly selected adult has an IQ scores that is greater than 140.




This is not a site that can teach you about an entire subject matter. We are here to help with individual problems. If I were you, I'd look at maybe Khan Academy or something of the like that can explain Normal Distributions to you.
 
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