Question

according to the records of a soft - drink company, the bottles for their one - liter - sized products contain an average (mean) of 1.015 liters of beverage, with a standard deviation of 0.14 liters. as part of routine quality assurance, a sample of 50 bottles has been taken. the sample mean amount of beverage in these 50 bottles was 0.99 liters. assuming the companys records are correct, find the probability of observing a sample mean of 0.99 liters or less in a sample of 50 bottles. carry your intermediate computations to at least four decimal places. round your answer to at least three decimal places.

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard - deviation and $n$ is the sample size. Given $\sigma = 0.14$ and $n = 50$, we have $\sigma_{\bar{x}}=\frac{0.14}{\sqrt{50}}\approx\frac{0.14}{7.0711}\approx0.0198$.

Step2: Calculate the z - score

The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean. Given $\bar{x}=0.99$, $\mu = 1.015$, and $\sigma_{\bar{x}}\approx0.0198$, we have $z=\frac{0.99 - 1.015}{0.0198}=\frac{- 0.025}{0.0198}\approx - 1.26$.

Step3: Find the probability

We want to find $P(\bar{X}\leq0.99)$, which is equivalent to $P(Z\leq - 1.26)$ using the standard normal distribution. Looking up the value in the standard - normal table, $P(Z\leq - 1.26)=0.104$.

Answer:

$0.104$