Math 011- MAC24 Name: Note, on Exam 2 you will be asked to do problems similar t
ID: 3050763 • Letter: M
Question
Math 011- MAC24 Name: Note, on Exam 2 you will be asked to do problems similar to these using the table of Normal probabiities (the a-table). it is good to know how to do problems like these unling the 'normalcdfl 'function on your calculator, but I will require you to show me on the exam that you can use the table also. The labels on boxes of Cheesy Poofs say that the box contains 16 ounces of Cheesy Poofs, but that doesn't guarantee that each box contains exactly 16.00 ounces. The machine that fills the boxes doesn't always put out the exact same amount. The amount in each box is Normally distributed, and has a standard deviation is.40 ounces. To help assure that most boxes contain as much or more than the label says, the machine is set so that the mean will be 16.28 ounces. What proportion of the boxes contain... 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? f ..between 15.80 and 16.00 ounces?Explanation / Answer
a) P(X < 16)
= P((X - mean)/sd < (16 - mean)/sd)
= P(Z < (16 - 16.28)/0.4)
= P(Z < -0.7)
= 0.2420
b) P(16 < X < 17)
= P((16 - 16.28)/0.4 < Z < (17 - 16.28)/0.4)
= P(-0.7 < Z < 1.8)
= P(Z < 1.8) - P(Z < -0.7)
= 0.9641 - 0.2420
= 0.7221
c) P(X > 16.5)
= P(Z > (16.5 - 16.28)/0.4)
= P(Z > 0.55)
= 1 - P(Z < 0.55)
= 1 - 0.7088 = 0.2912
d) P(16.5 < X < 16.9)
= P((16.5 - 16.28)/0.4 < Z < (16.9 - 16.28)/0.4)
= P(0.55 < Z < 1.55)
= P(Z < 1.55) - P(Z < 0.55)
= 0.9394 - 0.7088
= 0.2306
e) P(16.5 < X < 19)
= P((16.5 - 16.28)/0.4 < Z < (19 - 16.28)/0.4)
= P(0.55 < Z < 6.8)
= P(Z < 6.8) - P(Z < 0.55)
= 1 - 0.2912
= 0.7088
f) (15.8 < X < 16)
= P((15.8 - 16.28)/0.4 < Z < (16 - 16.28)/0.4)
= P(-1.2 < Z < -0.7)
= P(Z < -0.7) - P(Z < -1.2)
= 0.2420 - 0.1151
= 0.1269
i) P(X > 16) = 0.9
or, P(Z > (16 - mean)/0.4) = 0.9
or, P(Z < (16 - mean)/0.4) = 0.1
or, (16 - mean)/0.4 = -2.33
or, mean = 16 - 0.4 * (-2.33)
or, mean = 16.932
j) P(X > 16) = 0.9
or, P(Z > (16 - mean)/0.1) = 0.9
or, P(Z < (16 - mean)/0.1) = 0.1
or, (16 - mean)/0.1 = -2.33
or, mean = 16 - 0.1 * (-2.33)
or, mean = 16.233