Two independent simple random samples are taken to test the difference between the means of two populations whose variances are
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Answer:
f(x) = Three-halves (three-halves) Superscript x
f(x) = Two-thirds (three-halves) Superscript x
Step-by-step explanation:
Since, a function in the form of
Where, a and b are any constant,
is called exponential function,
There are two types of exponential function,
- Growth function : If b > 1,
- Decay function : if 0 < b < 1,
Since, In
Thus, it is a decay function.
in
Thus, it is a decay function.
in
Thus, it is a growth function.
in
Thus, it is a growth function.
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