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.

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

Step-by-step 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

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%