(t-distribution) A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. To maintain this average, 16 laptop batteries are tested. 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? Assume the distribution of battery lives to be approximately normal.

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.