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.

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m