| In 2011, the per capita consumption of chocolate in Switzerland was reported to be 20.63 pounds. Assume that the per capita consumption of chocolate in Switzerland is normally distributed with a mean of 20.63 pounds and a standard deviation of 6 pounds. | | | |
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| a. | What is the probability that someone in Switzerland consumed more than 12 pounds of chocolate in 2011? | | | |
| b. | What is the probability that someone in Switzerland consumed between 5 and 17 pounds of chocolate in 2011? | | | |
| c. | What is the proability that someone in Switzerland consumed less than 17 pounds of chocolate in 2011? | | | | |
| d. | 99% of the people in Switzerland consumed less than how many pounds of chocolate? | | | | | |
| e. | 95% of the people in Sweden consumed more than how many pounds of chocolate? | | | | | |
| f. | The probability is 50% that a person in Sweden consumed between these two amounts of coffee (symmetrically distributed around the mean)? |
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| In 2011, the per capita consumption of chocolate in Switzerland was reported to be 20.63 pounds. Assume that the per capita consumption of chocolate in Switzerland is normally distributed with a mean of 20.63 pounds and a standard deviation of 6 pounds. | | | |
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| | |
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| a. | What is the probability that someone in Switzerland consumed more than 12 pounds of chocolate in 2011? | | | |
| b. | What is the probability that someone in Switzerland consumed between 5 and 17 pounds of chocolate in 2011? | | | |
| c. | What is the proability that someone in Switzerland consumed less than 17 pounds of chocolate in 2011? | | | | |
| d. | 99% of the people in Switzerland consumed less than how many pounds of chocolate? | | | | | |
| e. | 95% of the people in Sweden consumed more than how many pounds of chocolate? | | | | | |
| f. | The probability is 50% that a person in Sweden consumed between these two amounts of coffee (symmetrically distributed around the mean)? |
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Normal Distribution
Mean ( u ) =20.63
Standard Deviation ( sd )=6
Normal Distribution = Z= X- u / sd ~ N(0,1)
a)
P(X > 12) = (12-20.63)/6
= -8.63/6 = -1.4383
= P ( Z >-1.438) From Standard Normal Table
= 0.9248
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 5) = (5-20.63)/6
= -15.63/6 = -2.605
= P ( Z <-2.605) From Standard Normal Table
= 0.00459
P(X < 17) = (17-20.63)/6
= -3.63/6 = -0.605
= P ( Z <-0.605) From Standard Normal Table
= 0.27259
P(5 < X < 17) = 0.27259-0.00459 = 0.268
c)
P(X < 17) = (17-20.63)/6
= -3.63/6= -0.605
= P ( Z <-0.605) From Standard Normal Table
= 0.2726
d)
P ( Z < x ) = 0.99
Value of z to the cumulative probability of 0.99 from normal table is 2.326
P( x-u/s.d < x - 20.63/6 ) = 0.99
That is, ( x - 20.63/6 ) = 2.33
--> x = 2.33 * 6 + 20.63 = 34.586
e)
P ( Z > x ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is -1.64
P( x-u/ (s.d) > x - 20.63/6) = 0.95
That is, ( x - 20.63/6) = -1.64
--> x = -1.64 * 6+20.63 = 10.76
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