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2 answers:
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Answer:
Grades between 62 and 64 result in a D grade.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, 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.
In this problem, we have that:
Find the numerical limits for a D grade.
D: Scores below the top 84% and above the bottom 10%
So below the 100-84 = 16th percentile and above the 10th percentile.
16th percentile:
This is the value of X when Z has a pvalue of 0.16. So X when Z = -0.995.
10th percentile:
This is the value of X when Z has a pvalue of 0.1. So X when Z = -1.28.
Grades between 62 and 64 result in a D grade.
Answer:
98% confidence interval for the average age of all students is [24.302 , 25.698]
Step-by-step explanation:
We are given that a random sample of 36 students at a community college showed an average age of 25 years.
Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.
So, the pivotal quantity for 98% confidence interval for the average age is given by;
P.Q. = ~ N(0,1)
where, = sample average age = 25 years
= population standard deviation = 1.8 years
n = sample of students = 36
= population average age
So, 98% confidence interval for the average age, is ;
P(-2.3263 < N(0,1) < 2.3263) = 0.98
P(-2.3263 < < 2.3263) = 0.98
P( <
<
) = 0.98
P( <
<
) = 0.98
98% confidence interval for = [
,
]
= [ ,
]
= [24.302 , 25.698]
Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].