A humidifier is set to maintain humidity at 40. Assuming humidity is distributed normally with the standard deviation of .1, what is the probability that humidity will be below 39.5?

Respuesta :

Answer:

0% probability that humidity will be below 39.5

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 = 40, \sigma = 39.5[/tex]

What is the probability that humidity will be below 39.5?

This is the pvalue of Z when X = 39.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{39.5 - 40}{0.1}[/tex]

[tex]Z = -5[/tex]

[tex]Z = -5[/tex] has a pvalue of 0.

0% probability that humidity will be below 39.5