Last year Lenny had an annual earned income of $58,475. He also had passive income of $1,255, and capital gains of $2,350. What
2 answers:
<u>Answer:</u> d. $62,080
<u>Step-by-step explanation:</u>
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The capital gain is the profit earned from an investment whereas the passive income is the income generated by very minimal daily efforts.
Given: Annual income earned by Lenny =
Passive income =
Capital gain =
Now,
Hence, Lenny's total gross income for the year = $62,080
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Step-by-step explanation:
Answer:
Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.
There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.
Step-by-step explanation:
Problems of normally distributed samples 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
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem
Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that .
What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?
For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:
has a pvalue of 0.9495.
This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.
For the average of the sample
What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?
Now, we need to find the standard deviation of the sample before using the zscore formula. That is:
.
Now
has a pvalue of 0.9999.
This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.