a team's stadium has a capacity of 86,047. The fan base is notorious for selling out of tickets every game. If every game sells
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Answer:
a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.
b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 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 373 minutes and standard deviation 67 minutes. So
A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.8907.
So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.
Lean
Normally distributed with mean 526 minutes and standard deviation 107 minutes. So
B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.0078.
So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.
Answer:
Probability that next week's show will have between 30 and 37 million viewers is 0.2248.
Step-by-step explanation:
We are given that the distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 26 million with a standard deviation of 8 million.
<em>Let X = number of viewers for the American Idol television show</em>
So, X ~ N()
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
where, = population mean = 26 million
= standard deviation = 8 million
So, probability that next week's show will have between 30 and 37 million viewers is given by = P(30 < X < 37) = P(X < 37) - P(X 30)
P(X < 37) = P( <
) = P(Z < 1.38) = 0.91621
P(X 30) = P(
) = P(Z
0.50) = 0.69146
<em>Therefore, P(30 < X < 37) = 0.91621 - 0.69146 = 0.2248</em>
Hence, probability that next week's show will have between 30 and 37 million viewers is 0.2248.
My key can’t perform this but I can solve it on a sheet...
So here I really hopes this helps
So here I really hopes this helps