the drama club is holding auditions for 5 different parts in a one-act play. if 9 people audition, how many ways can the roles b
1 answer:
Answer:
The roles can be assigned in 15120 ways.
Step-by-step explanation:
The parts, which means that the order in which the people are selected is important. So we use the permutations formula to solve this question.
Permutations formula:
The number of possible permutations of x elements from a set of n elements is given by the following formula:
In this question:
5 people chosen from a set of 9. So
The roles can be assigned in 15120 ways.
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Answer:
10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes
Step-by-step explanation:
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
.
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.
Mildly obese
Normally distributed with mean 375 minutes and standard deviation 68 minutes. So
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.8962.
So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
Lean
Normally distributed with mean 522 minutes and standard deviation 106 minutes. So
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.0045.
So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes