Respuesta :
Answer:
The probability that the mean of this sample is less than 16.1 ounces of beverage is 0.0537.
Step-by-step explanation:
We are given that the average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces.
A random sample of sixty-five 16-ounce beverage cans are selected
Let [tex]\bar X[/tex] = sample mean amount of a beverage
The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean amount of a beverage = 16.18 ounces
[tex]\sigma[/tex] = standard deviation = 0.4 ounces
n = sample of 16-ounce beverage cans = 65
Now, the probability that the mean of this sample is less than 16.1 ounces of beverage is given by = P([tex]\bar X[/tex] < 16.1 ounces)
P([tex]\bar X[/tex] < 16.1 ounces) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{16.1-16.18}{\frac{0.4}{\sqrt{65} } }[/tex] ) = P(Z < -1.61) = 1 - P(Z [tex]\leq[/tex] 1.61)
= 1 - 0.9463 = 0.0537
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9591.
