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. If a random sample of sixty-five 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 16.1 ounces of beverage? Answer: (keep 4 decimal places)
Solution :
Given that ,
mean = = 16.18
standard deviation = = 0.4
n = 65
_{} = = 16.18
_{} = / n = 0.4 / 65 = 0.0496
P( < 16.1) = P(( - _{} ) / _{} < (16.1 - 16.18) / 0.0496)
= P(z < -1.61)
Using z table
= 0.0537
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