Find the sum of each series, if it exists91 + 85 + 79 + … + (29)
1 answer:
Answer:
651.
Step-by-step explanation:
Note: In the given series it should be -29 instead of 29 because 29 cannot be a term of AP whose first term is 91 and common difference is -6.
Consider the given series is
It is the sum of an AP. Here,
First term = 91
Common difference = 85 - 91 = -6
Last term = -29
nth term of an AP is
where, a is first term and d is common difference.
Sum of AP is
Therefore, the sum of given series is 651.
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Answer:
P(X > 25) = 0.69
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
The sale prices for a particular car are normally distributed with a mean and standard deviation of 26 thousand dollars and 2 thousand dollars, respectively.
This means that
Find P(X>25)
This is 1 subtracted by the pvalue of Z when X = 25. So
has a pvalue of 0.31
1 - 0.31 = 0.69.
So
P(X > 25) = 0.69
Answer:
If we compare the p value and a significance level for example we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.
Step-by-step explanation:
Data given and notation
represent the average lateral recumbency for the sample
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to apply a left tailed test.
What are H0 and Ha for this study?
Null hypothesis:
Alternative hypothesis :
Compute the test statistic
The statistic for this case is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
The degrees of freedom are given by:
Give the appropriate conclusion for the test
Since is a one side left tailed test the p value would be:
Conclusion
If we compare the p value and a significance level for example we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.