Respuesta :
Answer:
95.99% probability that a randomly selected game will be completed in 200 minutes or less.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
In this problem, we have that:
[tex]\mu = 172, \sigma = 16[/tex]
What is the probability that a randomly selected game will be completed in 200 minutes or less
This is the pvalue of Z when X = 200. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200 - 172}{16}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a pvalue of 0.9599
95.99% probability that a randomly selected game will be completed in 200 minutes or less.
Answer:
Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.
Step-by-step explanation:
We are given that during a recent season, games averaged 2 hours and 52 minutes (172 minutes) to complete. Assume the length of games follows the normal distribution with a standard deviation of 16 minutes.
Let X = length of games
So, X ~ N([tex]\mu = 172, \sigma^{2} = 16^{2})[/tex]
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = average time = 172 minutes
[tex]\sigma[/tex] = standard deviation = 16 minutes
So, Probability that a randomly selected game will be completed in 200 minutes or less is given by = P(X [tex]\leq[/tex] 200 min)
P(X [tex]\leq[/tex] 200) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{200-172}{16}[/tex] ) = P(Z [tex]\leq[/tex] 1.75)
= 0.95994
Hence, Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.
