A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.2 in. A branch of the m
1 answer:
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Answer:
B. (1,7)
Step-by-step explanation:
We can substitute the x and y values of each coordinate into the inequality and test if they work.
Let's start with A, 5 being y and 0 being x .
5 IS NOT greater than 5, they are the exact same, so A is out.
Let's try B, 1 being x and 7 being y.
7 IS greater than 6, so B. (1,7) does work for this inequality!
Let's do C for fun, when 7 is x and 1 is y.
1 IS NOT greater than 12, it is quite less than 12, so C doesn't work.
Therefore B. (1,7) works for the inequality of .
Hope this helped!
Answer:
The percentle for Abby's score was the 89.62nd percentile.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation(which is the square root of the variance)
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. 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 pvalue, we get the probability that the value of the measure is greater than X.
Abby's mom score:
93rd percentile in the math SAT exam. In 1982 the mean score was 503 and the variance of the scores was 9604.
93rd percentile. X when Z has a pvalue of 0.93. So X when Z = 1.476.
So
Abby's score
She scored 648.
So
has a pvalue of 0.8962.
The percentle for Abby's score was the 89.62nd percentile.