Question 1199684: Problem B:
The incubation time for chicks is normally distributed with a mean of 21 days, and a standard deviation of approximately 1.2 days (based on the information from the World Bank Encyclopedia). If 1000 eggs are being incubated, how many chicks do we expect will hatch
c.) In 18.3 days or fewer?
d.) In at least 22.8 days?
Answer by ikleyn(52775)   (Show Source):
You can put this solution on YOUR website! .
Problem B:
The incubation time for chicks is normally distributed with a mean of 21 days,
and a standard deviation of approximately 1.2 days
(based on the information from the World Bank Encyclopedia).
If 1000 eggs are being incubated, how many chicks do we expect will hatch
(c) In 18.3 days or fewer?
(d) In at least 22.8 days?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(c) A normal distribution curve is a bell shaped curve.
The total area under each such curve is 1.
Of the 1000 eggs, the number of chicks that are expected to hatch,
in part (c) make a ratio, which is the area under this specified normal curve
on the  of the raw mark z= 18.3 days.
THEREFORE, our first step is to calculate the area under given normal curve
on the left of the raw mark z = 18.3 days.
You can do it using the normal cumulative distribution function mormalcdf.
z1 z2 mu SD <<<---=== formatting
In your calculator it is p = normalcdf(-9999, 18.3, 21, 1.2) = 0.0122.
After getting this number p = 0.0122 (the probability, or the area under the curve),
you multiply 1000 by p, and you get the desired ANSWER
+--------------------------------------------------------------------------+
| the number of chicks that are expected to hatch is 1000*0.0122 = 12.2, |
| which we round to the closest integer number 12. |
+--------------------------------------------------------------------------+
(d) Similarly, in part (d), our first step is to calculate the area under given normal curve
on the  of the raw mark z = 22.8 days.
You can do it using the normal cumulative distribution function mormalcdf.
z1 z2 mu SD <<<---=== formatting
In your calculator it is p = normalcdf(22.8, 9999, 21, 1.2) = 0.0668.
After getting this number p = 0.0668 (the probability, or the area under the curve),
you multiply 1000 by p, and you get the desired ANSWER
+---------------------------------------------------------------------------+
| the number of chicks that are expected to hatch is 1000*0.0668 = 66.8, |
| which we round to the closest integer number 67. |
+---------------------------------------------------------------------------+
Solved.
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