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Math 011- MAC22 Name: to do problems similar to these using the table of Normal

ID: 3074503 • Letter: M

Question

Math 011- MAC22 Name: to do problems similar to these using the table of Normal probabilities (the -table). Itis like these using the 'normaledff function on your calculator, but I will require you to Note, on Exam 3 you will be asked problems good to know how to do problems using t show me on the exam that you can use the table also. but that doesn't The labels on boxes of Cheesy Poofs say that the box contains 16 ounces of Cheesy Poofs, guarantee that each box contains exactly 16.00 ounces. The machine that fills the box es doesn't always put out the exact deviation is.40 ounces. T machine is set so that the mean will be 16.28 ounces. What proportion of the boxes contain... same amount. The amount in each box is Normally distributed, and has a standard o help assure that most boxes contain as much or more than the label says, the a) ...less than 16.00 ounces? a) b).between 16.00 and 17.00 ounces? b) c) ..more than 16.50 ounces? c) d) ...between 16.50 and 16.90 ounces? d) e).between 16.50 and 19.00 ounces? e) f between 15.80 and 16.00 ounces?

Explanation / Answer

Solution:

Here, we are given that the variable follows a normal distribution.

Mean = 16.28

SD = 0.40

Part a

Here, we have to find P(X<16)

Z = (X – mean) / SD

Z = (16 – 16.28) / 0.40

Z = -0.28/0.40

Z = -0.7

P(Z<-0.7) = 0.241964

(By using z-table or calculator or excel)

P(X<16) = 0.241964

Answer: 0.241964

Part b

Here, we have to find P(16<X<17)

P(16<X<17) = P(X<17) – P(X<16)

First calculate P(X<17)

Z = (17 - 16.28)/0.40

Z = 1.8

P(Z<1.8) = 0.96407

(By using z-table or calculator or excel)

P(X<17) = 0.96407

Now calculate P(X<16)

Z = (16 - 16.28)/ 0.40

Z = -0.7

P(Z<-0.7) = 0.241964

(By using z-table or calculator or excel)

P(X<16) = 0.241964

P(16<X<17) = P(X<17) – P(X<16)

P(16<X<17) = 0.96407 - 0.241964

P(16<X<17) = 0.722106

Required probability or proportion = 0.722106

Part c

Here, we have to find P(X>16.50)

P(X>16.50) = 1 – P(X<16.50)

Z = (16.50 - 16.28) / 0.40

Z = 0.55

P(Z<0.55) = 0.70884

(By using z-table or calculator or excel)

P(X<16.50) = 0.70884

P(X>16.50) = 1 – P(X<16.50)

P(X>16.50) = 1 – 0.70884

P(X>16.50) = 0.29116

Required probability or proportion = 0.29116

Part d

Here, we have to find P(16.50<X<16.90)

P(16.50<X<16.90) = P(X<16.90) – P(X<16.50)

First calculate P(X<16.90)

Z = (16.90 – 16.28) / 0.40

Z = 1.55

P(Z<1.55) = 0.939429

(By using z-table or calculator or excel)

P(X<16.9) = 0.939429

Now, calculate P(X<16.50)

Z = (16.50 - 16.28) / 0.40

Z = 0.55

P(Z<0.55) = 0.70884

(By using z-table or calculator or excel)

P(X<16.50) = 0.70884

P(16.50<X<16.90) = P(X<16.90) – P(X<16.50)

P(16.50<X<16.90) = 0.939429 - 0.70884

P(16.50<X<16.90) = 0.230589

Required proportion = 0.230589