Least to greatest 7/8, 6/13, and 5/9
1 answer:
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Answer:
66.98% probability that the mean gain for the sample was between 250 and 500.
Step-by-step explanation:
To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.
Normal probability distribution
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:
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 random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
In this problem, we have that:
What is the probability that the mean gain for the sample was between 250 and 500?
This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.
So
X = 500
By the Central Limit Theorem
has a pvalue of 0.7257.
X = 250
By the Central Limit Theorem
has a pvalue of 0.0559.
So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.
You need to find "two-fifths of 30." Of here means multiplication:
There are 12 athletes in the club.