# Normal Distribution and Probabilities in Industries

Can you please assist with questions 6-11 and 6-17 putting the answers and solutions in an excel spread sheet.

6.11 ) Consider a random variable, z, that has a standardized normal distribution. Determine the following probabilities:

a) P(0<z<1.96)

b) P(z>1.645)

c) P(1.28<z<2.33)

d) P(-2<z<3)

e) P(z>-1)

6.17 ) Bowser Bites Industries (BBI) sells large bags of dog food to warehouse clubs. BBI uses an automatic filling process to fill the bags. Weights of the filled bag are approximately normally distributed with a mean of 50 kilograms and standard deviation of 1.25 kilograms.

a) What is the probability that a filled bag will weigh less than 49.5 kilograms?

b) What is the probability that a randomly sampled filled bag will weigh between 48.5 and 51 kilograms?

c) What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled?

d) BBI is unable to adjust the mean of the filling process. However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that no more than 2% of all filled bags weigh more than 52 kilograms?

See the attached file.

© BrainMass Inc. brainmass.com July 23, 2018, 1:51 pm ad1c9bdddf#### Solution Preview

Please see the attached Excel file:

6.11 ) Consider a random variable, z, that has a standardized normal distribution. Determine the following probabilities:

We will use EXCEL worksheet function NORMSDIST to calculate the probabilities

a) P(0<z<1.96)

Between z= 0 and 1.96

z1= 0

z2= 1.96

Cumulative Probability corresponding to z1= 0 is= 0.5 =NORMSDIST(0) 0r= 50.00%

Cumulative Probability corresponding to z2= 1.96 is= 0.975 =NORMSDIST(1.96) 0r= 97.50%

Therefore probability that the value of z will be between z1= 0.00 and z2= 1.96

is = 47.50% =97.5%-50.%

Answer: 47.50% or 0.475

b) P(z>1.645)

To the right of z= 1.645

z= 1.645

Cumulative probability corresponding to Z= 1.645 is 0.95 =NORMSDIST(1.645) 0r= 95.00%

Thus area to the left of z= 1.645 is 95.00%

Hence area to the right of z= 1.645 is 5.00% =100% - 95.%

Answer: 5.00% 0r 0.0500

c) P(1.28<z<2.33)

Between z= 1.28 and 2.33

z1= 1.28

z2= 2.33

Cumulative Probability corresponding to z1= 1.28 is= 0.8997 =NORMSDIST(1.28) 0r= 89.97%

Cumulative ...

#### Solution Summary

Uses EXCEL worksheet functions NORMSDIST, NORMSINV to calculate normal distribution probabilities.