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Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 1.2 ounces. Round your answers to 4 decimal places.

(a) If one potato is randomly selected, find the probability that it weighs less than 10 ounces.


(b) If one potato is randomly selected, find the probability that it weighs more than 12 ounces.


(c) If one potato is randomly selected, find the probability that it weighs between 10 and 12 ounces.

User Koen Weyn
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2 Answers

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Answer:

(a): 0.7967

(b): 0.0062

(c): 0.1771

Explanation:

Solution :

Given that ,

mean = \mu = 9

standard deviation = \sigma = 1.2

(a)

P(x < 10) = P[(x - \mu ) / \sigma < (10 - 9) / 1.2]

= P(z < 0.83)

= 0.7967

Probability = 0.7967

(b)

P(x > 12) = 1 - P(x < 12)

= 1 - P[(x - \mu ) / \sigma < (12 - 9) / 1.2]

= 1 - P(z < 2.5)

= 1 - 0.9938

= 0.0062

Probability = 0.0062

(c)

P(10 < x < 12) = P[(10 - 9)/ 1.2) < (x - \mu ) /\sigma < (12 - 9) / 1.2) ]

= P(0.83 < z < 2.5)

= P(z < 2.5) - P(z < 0.83)

= 0.9938 - 0.7967

= 0.1971

Probability = 0.1771

User Alfredo Luco G
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4 votes

Answer: a) 79.10 % b) 7% c)19.85 %

Explanation:

a) Z = ( × - μ ) ÷ σ Where μ mean of population σ standard deviation Z is the abscissa to give the area or probability we are looking for associated to the value 10 ounces ( × )

So: Z = ( 10 - 9 ) ÷ 1.2 ⇒ Z = 1/1.2 ⇒ Z = 0.83 it has to be below this value

From Z tables we get P [ Z ≤ 0.82 ] = 0.7910 0r 79.10 %

b) Following the same procedure: We look for

P [ Z > ( x - μ ) ÷ σ ] ⇒ Z = ( 12 -10 ) ÷ 1.2 = 2.5

From Z table we get the area under the curve from the left tail up to the point Z < 2.5 ( 2.5 not included) but we were asked for the area out of that previous so 1- 0.9930 = 0.007 is the area we are looking for

So P (b) = 0.007 or 7 %

Finally between the two points above mentioned ( 10 ≤ Z ≤ 12 ) we use the previous values (taking in consideration the limits, according to the problem statement )

Z ≤ 10 Z ⇒( 10-9 ) ÷ 1.2 Z = 0.7967

Z ≥ 12 Z ⇒ ( 12 - 9 ) ÷ 1.2 Z = 0.9952

The interval is between these two points

0.9952 - 0.7967 = 0.1985 ⇒ or 19.85 %

The attached help in the understanding of the solution

Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with-example-1
Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with-example-2
User BitsAreNumbersToo
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