Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of 0.24 ounces. The weights of the sugar bags are normally distributed. What is the probability that 16 randomly selected packages will have an average weight less than 15.97 ounces?

the probability of less than 15.97 will be (16-7) / 16 = 0.5625

Step-by-step explanation:

To ascertain the manager claim, we use the normal distribution curve.

z = (x - x') / σ

, where

z is called the normal standard variate,

x is the value of the variable, = 15.97

x' is the mean value of the distribution = 16

σ is the standard deviation of the distribution = 0.24

so, z = (15.97 - 16) / 0.24 = - 0.125

Using a table of normal distribution to check the partial areas beneath the standardized normal curve, a z-value of - 0.125 corresponds to an area of 0.0498 between the mean value. The negative z-value shows that it lies to the left of the z=0 ordinate.

The total area under the standardized normal curve is unity and since the curve is symmetrical, it follows that the total area to the left of the z=0 ordinate is 0.5000. Thus the area to the left of the z=-0.125 ordinate is 0.5000-0.0498 = 0.4503 of the total area of the curve.

therefore, the probability of the average weight equal to 15.97 ounces is 0.4503.

For a group of 16 packages, 16*0.4503 = 7.2, i. e. 7 packages will have the weight of 15.97 ounces.

the probability of less than 15.97 will be (16-7) / 16 = 0.5625