Respuesta :
Answer:
a) 0.9738 = 97.38% probability that daily production is less than 36.4 liters.
b) 0.7673 = 76.73% probability that daily production is more than 27.6 liters
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.
In this problem, we have that:
[tex]\mu = 30, \sigma = 3.3[/tex]
A) What is the probability that daily production is less than 36.4 liters?
This is the pvalue of Z when X = 36.4. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{36.4 - 30}{3.3}[/tex]
[tex]Z = 1.94[/tex]
[tex]Z = 1.94[/tex] has a pvalue of 0.9738.
0.9738 = 97.38% probability that daily production is less than 36.4 liters.
B) What is the probability that daily production is more than 27.6 liters?
This is 1 subtracted by the pvalue of Z when X = 27.6. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{27.6 - 30}{3.3}[/tex]
[tex]Z = -0.73[/tex]
[tex]Z = -0.73[/tex] has a pvalue of 0.2327.
1 - 0.2327 = 0.7673
0.7673 = 76.73% probability that daily production is more than 27.6 liters
