The grade point average in an econometrics examination was normally distributed with a mean of 75. in a sample of 10 percent of
1 answer:
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Answer:
Sebastian didn't interpret the numbers correctly (See explanation)
Step-by-step explanation:
If you go up an elevator, you are gaining floors, so the number is positive.
If you go down an elevator, you are losing floors, so the number is positive.
Looking at the statement, he went 7 floors up to his room, then 9 floors down to the parking garage this can be represented as , since the 7 is positive and the 9 is negative.
Sebastian described the movement as , which is wrong - 9 should be -9 and -7 should be 7.
Hope this helped!
Answer:
a) P(12.99 ≤ X ≤ 13.01) = 0.3840
b) P(X ≥ 13.01) = 0.3075
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the cental limit theorem.
Normal probability distribution
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.
Here we have
This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.
X = 13.01
By the Central Limit Theorem
has a pvalue of 0.6915
X = 12.99
has a pvalue of 0.3075
0.6915 - 0.3075 = 0.3840
P(12.99 ≤ X ≤ 13.01) = 0.3840
(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?
P(X ≥ 13.01) =
This is 1 subtracted by the pvalue of Z when X = 13.01. So
has a pvalue of 0.6915
1 - 0.6915 = 0.3075
P(X ≥ 13.01) = 0.3075