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? (round to 4 decimal places)

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 = sample mean amount of a beverage

The z-score probability distribution for the sample mean is given by;

                          Z  =    ~ N(0,1)

where, = population mean amount of a beverage = 16.18 ounces

            = 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( < 16.1 ounces)

   P( < 16.1 ounces) = P( < ) = P(Z < -1.61) = 1 - P(Z 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.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)