Raw scores on standardized tests are often transformed for easier comparison. A test of reading ability has a mean of 75 and a s tandard deviation of 10 when given to third-graders. Sixth-graders have a mean score of 82 and a standard deviation of 11 on the same test. What percent of sixth-graders earn a score of at least 93 on the reading test?
1 answer:
Using the normal distribution , it is found that 3.59% of sixth-graders earn a score of at least 93 on the reading test.
-------------------------
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which is given by:
is the mean . is the standard deviation. It measures how many standard deviations the measure is from the mean. Each z-score has a p-value associated with it. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
On the reading test, the mean is of 75 , thus . The standard deviation is of 10 , thus . The proportion who scored above 93 is <u>1 subtracted by the p-value of Z when X = 93</u>, thus:
has a p-value of 0.9641.
1 - 0.9641 = 0.0359.
0.0359 x 100% = 3.59%
3.59% of sixth-graders earn a score of at least 93 on the reading test.
A similar problem is given at brainly.com/question/24663213
You might be interested in
Answer:
123
Step-by-step explanation:
180-57=123
OK OK ok
Answer:
10
Step-by-step explanation:
14 - (17 - 5) + 8
PEMDAS says parentheses first
14 - (12) + 8
Since we have no exponents, or multiplication or division
We add and subtract from left to right.
2+8
10
Answer:
x=5
Step-by-step explanation:
7x-17=18
7x=18+17
7x=35
x=35/7
x=5
X = 6 and -6
if you need anything explained, please ask
Here you go. I hope you understand. Ask if you don't.