the distribution of weight for 9 - ounce bags of a particular brand of potato chips can be modeled by a…

the distribution of weight for 9 - ounce bags of a particular brand of potato chips can be modeled by a normal distribution with mean $mu = 9.12$ ounces and standard deviation $sigma=0.05$ ounce. sketch the normal density curve. label the mean and the points that are 1, 2, and 3 standard deviations from the mean. do not round your answers.

the distribution of weight for 9 - ounce bags of a particular brand of potato chips can be modeled by a normal distribution with mean $mu = 9.12$ ounces and standard deviation $sigma=0.05$ ounce. sketch the normal density curve. label the mean and the points that are 1, 2, and 3 standard deviations from the mean. do not round your answers.

Answer

Explanation:

Step1: Calculate 1 - standard - deviation points

The formula to find the points that are (k) standard deviations from the mean (\mu) is (x=\mu\pm k\sigma). For (k = 1), (x_1=\mu-\sigma=9.12 - 0.05=9.07) and (x_2=\mu+\sigma=9.12 + 0.05=9.17)

Step2: Calculate 2 - standard - deviation points

For (k = 2), (x_3=\mu-2\sigma=9.12-2\times0.05=9.12 - 0.10 = 9.02) and (x_4=\mu + 2\sigma=9.12+2\times0.05=9.12 + 0.10=9.22)

Step3: Calculate 3 - standard - deviation points

For (k = 3), (x_5=\mu-3\sigma=9.12-3\times0.05=9.12 - 0.15=8.97) and (x_6=\mu+3\sigma=9.12+3\times0.05=9.12 + 0.15=9.27) The mean is (\mu = 9.12).

Answer:

Label the mean as (9.12). Label the points: (8.97) (3 standard - deviations below the mean), (9.02) (2 standard - deviations below the mean), (9.07) (1 standard - deviation below the mean), (9.17) (1 standard - deviation above the mean), (9.22) (2 standard - deviations above the mean), (9.27) (3 standard - deviations above the mean)