Answer:
0.3075 = 30.75% probability that a person will wait for more than 7 minutes.
Step-by-step explanation:
Problems of normally distributed samples are 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.
The standard deviation is the square root of the variance.
In this problem, we have that:
[tex]\mu = 6, \sigma = \sqrt{4} = 2[/tex]
Find the probability that a person will wait for more than 7 minutes.
This is 1 subtracted by the pvalue of Z when X = 7. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7 - 6}{2}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915
1 - 0.6915 = 0.3075
0.3075 = 30.75% probability that a person will wait for more than 7 minutes.
The probability that a person will wait for more than 7 minutes is 30.85%.
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 6 minutes and variance = 4 minutes, hence:
Standard deviation = √variance = √4 = 2 minutes
For > 7 minutes:
z = (7 - 6)/2 = 0.5
P(z > 0.5) = 1 - P(z < 0.5) = 1 - 0.6915 = 0.3085
The probability that a person will wait for more than 7 minutes is 30.85%.
Find out more on z score at: https://brainly.com/question/25638875
