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The amount of detergent dispensed into bottles of liquid laundry detergent bottles for a particular brand is normally distributed with a mean of 84.5 ounces with a standard deviation of 1.1 ounces. If seventeen bottles are randomly chosen from the factory, what is the probability that the mean fill is more than 84.8 ounces

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The probability that the mean fill is more than 84.8 ounces is 0.39358

How to determine the probability that the mean fill is more than 84.8 ounces?

From the question, the given parameters about the distribution are

  • Mean value of the set of data = 84.5
  • Standard deviation value of the set of data = 1.1
  • The actual data value = 84.8

The z-score of the data value is calculated using the following formula

z = (x - mean value)/standard deviation

Substitute the given parameters in the above equation

z = (84.8 - 84.5)/1.1

Evaluate the difference of 84.8 and 84.5

z = 0.3/1.1

Evaluate the quotient of 0.3 and 1.1

z = 0.27

The probability that the mean fill is more than 84.8 ounces is then calculated as:

P(x > 84.8) = P(z > 0.27)

From the z table of probabilities, we have;

P(x > 84.8) = 0.39358

Hence, the probability that the mean fill is more than 84.8 ounces is 0.39358

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