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Normal Distribution and Probabilities in Industries

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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.

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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.

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Quantitative Analysis 3 multiple choice questions: probability, Poisson distribution, Normal Distribution

1. Meagan Dubean manages a portfolio of 200 common stocks. Her staff classified the portfolio stocks by 'industry sector' and 'investment objective.'

Investment objective Industry Sector
Electronics Airlines Healthcare Total
Growth 84 21 35 140
Income 36 9 15 60
Total 120 30 50 200

Are "Healthcare" and "Income" independent and why or why not?

a. yes, because P(Income intersection Healthcare) 0
b. no, because P(Income intersection Healthcare) = P(Healthcare) ? P(Income)
c. yes, because P(Healthcare | Income) = P(Healthcare)
d. no, because P(Healthcare) not equal to P(Income)

2. If lambda is 3 occurrences per five minute time interval, the probability of getting 5 occurrences over a five minute interval is _______.
a. 0.1500
b. 0.0940
c. 0.1008
d. 0.0417

3. Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers. Accordingly, his staff recorded the waiting times for 64 randomly selected walk-in customers, and determined that their mean waiting time was 15 minutes and that the standard deviation was 4 minutes. The 95% confidence interval for the population mean of waiting times is ________.

a. 7.16 to 22.84
b. 14.06 to 15.94
c. 8.42 to 21.58
d. 13.88 to 16.12

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