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Answer:
And if we solve for a we got;
So the value of height that separates the bottom 95% of data from the top 5% is 11.64
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent certain lifetime of a population, and for this case we know the distribution for X is given by:
Where and
For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got;
So the value of height that separates the bottom 95% of data from the top 5% is 11.64
Answer:
P-value (t=2.58) = 0.0066.
Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean μ exceeds 2.4.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=45.
The sample mean is M=2.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.