Question

A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a normal distribution with mean 16.05 ounces with a standard deviation of .1 ounces. If four bottles are randomly selected each hour and the number of ounces in each bottle is measured, then 95% of the means calculated should occur in what interval? Hint: the standard deviation rule says that 95% of the observations are within how many standard deviations away from the mean? Round answers to four decimal places

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Answer 1

Solution:-

Given that,

mean = = 16.05

standard deviation = = 0.1

n = 4

= = 16.05

= / n = 0.1/ 4 = 0.05

Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula  

= z * +

= -1.96 * 0.05 + 16.05

= 15.952 ounces

Using z-score formula  

= z * +

= 1.96 * 0.05 + 16.05

= 16.148 ounces

95% interval is = 15.952 and 16.148 ounces

±1.96 standard deviation away from mean