Respuesta :
Answer:
0.6832 = 68.32% probability that a given student will complete the test in more than 35 minutes but less than 43 minutes
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 41.0 minutes and a variance of 3.4 minutes.
This means that [tex]\mu = 41, \sigma = 3.4[/tex]
What is the probability, rounded to four decimal places, that a given student will complete the test in more than 35 minutes but less than 43 minutes?
This is the p-value of Z when X = 43 subtracted by the p-value of Z when X = 35.
X = 43
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{43 - 41}{3.4}[/tex]
[tex]Z = 0.59[/tex]
[tex]Z = 0.59[/tex] has a p-value of 0.7224
X = 35
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 41}{3.4}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a p-value of 0.0392
0.7224 - 0.0392 = 0.6832
0.6832 = 68.32% probability that a given student will complete the test in more than 35 minutes but less than 43 minutes
