Respuesta :
Answer:
69.15% probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces.
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:
Mean of four ounces, so [tex]\mu = 4[/tex]
Variance of .25 squared ounces. The standard deviation is the square root of the variance, so [tex]\sigma = \sqrt{0.25} = 0.5[/tex]
What is the probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces?
This is 1 subtracted by the pvalue of Z when X = 3.75. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.75 - 4}{0.5}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085.
So there is a 1-0.3085 = 0.6915 = 69.15% probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces.
