Please answer this im dying here
2 answers:
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Using the normal distribution, it is found that:
a) 0.8599 = 85.99% probability that x is more than 60.
b) 0.1788 = 17.88% probability that x is less than 110.
c) 0.6811 = 68.11% probability that x is between 60 and 110.
d) 0.0643 = 6.43% probability that x is greater than 125.
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
In this problem:
- The mean is of 87, thus
.
- The standard deviation is of 25, thus
.
Item a:
This probability is <u>1 subtracted by the p-value of Z when X = 60</u>, thus:
has a p-value of 0.1401.
1 - 0.1401 = 0.8599
0.8599 = 85.99% probability that x is more than 60.
Item b:
This probability is the <u>p-value of Z when X = 110</u>, thus:
has a p-value of 0.8212.
1 - 0.8212 = 0.1788.
0.1788 = 17.88% probability that x is less than 110.
Item c:
This probability is the <u>p-value of Z when X = 110 subtracted by the p-value of Z when X = 60</u>.
From the previous two items, 0.8212 - 0.1401 = 0.6811.
0.6811 = 68.11% probability that x is between 60 and 110.
Item d:
This probability is <u>1 subtracted by the p-value of Z when X = 125</u>, thus:
has a p-value of 0.9357.
1 - 0.9357 = 0.0643.
0.0643 = 6.43% probability that x is greater than 125.
A similar problem is given at brainly.com/question/24863330
Ruth Bader Ginsburg began her career as a justice where she left off as an advocate, fighting for women's rights. In 1996, Ginsburg wrote the majority opinion in United States v. Virginia, holding that qualified women could not be denied admission to Virginia Military Institute.
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