Respuesta :
Answer:
There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.
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:
A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. This means that [tex]\mu = 50[/tex].
We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter?
We have to find the pvalue of Z when [tex]X = 47.3[/tex].
We are working with a sample mean, so we use the standard deviation of the sample in the place of [tex]\sigma[/tex]. That is [tex]s = 9[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{47.3-50}{9}[/tex]
[tex]Z = -0.3[/tex]
[tex]Z = -0.3[/tex] has a pvalue of 0.3821.
There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.
