Respuesta :
Answer:
The minimum value of the bill that is greater than 95% of the bills is $37.87.
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:
[tex]\mu = 28, \sigma = 6[/tex]
What are the minimum value of the bill that is greater than 95% of the bills?
This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 28}{6}[/tex]
[tex]X - 28 = 6*1.645[/tex]
[tex]X = 37.87[/tex]
The minimum value of the bill that is greater than 95% of the bills is $37.87.
