Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviatio
n 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) 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.
1 answer:
Answer: 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
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The box and whisker plot is attached.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.