17. It takes one minute to fill (3/7) th of a vessel. What is the time taken in
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Answer:
The 90th percentile for the distribution of the total contributions is $6,342,525.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central 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.
For sums of size n, the mean is and the standard deviation is
In this question:
The 90th percentile for the distribution of the total contributions
This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then
By the Central Limit Theorem
The 90th percentile for the distribution of the total contributions is $6,342,525.
<h2>Answer:</h2>
A.
Let Annie's weight be = a
Let Benjie's weighs = b
Let Carmen's weight be = c
One day Annie weighed 24 ounces more than Benjie, equation forms:
......(1)
Benjie weighed 3 1/4 pounds less than Carmen.
In ounces:
1 pound = 16 ounces
pounds =
ounces
or
......(2)
Now adding (1) and (2), we get
a+b=b+24+c-52
=>
This gives Annie weighs 28 ounces less than Carmen.
B.
We cannot know anyone's actual weight, as we only know their relative weights.