Question 1202163: The incubation time for chicks is normally distributed with mean 21 days and standard deviation approximately 1 day. If 1000 eggs are being incubated, how many chicks do we expect will hatch?
a) in 19 to 23 days?
b) in 22 days or more?
c) in 21 days or fewer?
d) in 18 to 21 days?
e) in 19 days or fewer
f) in 19 to 24 days?
g) in 18 to 24 days?
h) in 18 days or more?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'll do parts (a), (b), and (c) to get you started.
Part (a)
mu = 21 = population mean
sigma = 1 = population standard deviation
x = incubation time in days
Find the z score when x = 19
z = (x - mu)/sigma
z = (19 - 21)/1
z = -2
Repeat for x = 23 to get z = 2
The task of finding P(19 < x < 23) is the same as P(-2 < z < 2).
Now use a table such as this one
https://www.ztable.net/
Similar tables are found in the back of your stats textbook.
An alternative is to use a stats calculator such as this
https://davidmlane.com/normal.html
However, I'll stick to using the table.
That table says:
P(Z < -2) = 0.02275
and
P(Z < 2) = 0.97725
Those values are approximate.
This gives
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)
P(-2 < Z < 2) = 0.97725 - 0.02275
P(-2 < Z < 2) = 0.9545
This value is approximate.
About 95.45% of the population is between the z scores z = -2 and z = 2.
This translates to roughly 95.45% of the population of eggs with incubation times between x = 19 and x = 23 days.
If there are 1000 eggs being incubated, then about 0.9545*1000 = 954.5 = 955 eggs will hatch in the timeframe of 19 days to 23 days.
Note that the Empirical Rule says roughly 95% of the population is within 2 standard deviations.
Answer: 955
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Part (b)
Find the z score when x = 22
z = (x - mu)/sigma
z = (22 - 21)/1
z = 1
We are 1 standard deviation above the mean.
Use the z table to find that
P(Z < 1.00) = 0.84134
which must mean
P(Z > 1.00) = 1-P(Z < 1.00)
P(Z > 1.00) = 1-0.84134
P(Z > 1.00) = 0.15866
About 15.866% of the population has an incubation time of 22 days or more.
15.866% of 1000 = 0.15866*1000 = 158.66 = 159
Answer: 159
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Part (c)
We could convert to a z score and use a table, but that might be a bit overkill here.
50% of the population is always below the mean of a normal distribution.
50% of 1000 = 0.50*1000 = 500
Answer: 500
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