Respuesta :
Answer:
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Step-by-step explanation:
Z-score
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.
What is the probability that a randomly selected truck from the fleet will have to be inspected?
Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.
Z = 2.5 has a pvalue of 0.9938
1 - 0.9938 = 0.0062
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Answer:
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the mileage of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
