The mean incubation time of fertilized chicken eggs kept at 100.5 degree F in a still-air incubator is 21 days. Suppose that the incubation times are approx normally distributed with a standard deviation of 1 day. Source: University of Illinois Extension (a) What is the probability that a randomly selected fertilized chicken egg hatches in less than 20 days? (b) What is the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch? (c) What is the probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

a) We want to find out the probability that a randomly selected fertilized chicken egg hatches in less than 20 days.

[tex]P(X < 20) = P(Z < \frac{x - \mu}{\sigma} )\\\\P(X < 20) = P(Z < \frac{20 - 21}{1} )\\\\P(X < 20) = P(Z < \frac{-1}{1} )\\\\P(X < 20) = P(Z < -1)\\[/tex]

The z-score corresponding to -1 is 0.1587

[tex]P(X < 20) = 0.1587\\\\P(X < 20) = 15.87 \%[/tex]

Therefore, the probability that a randomly selected fertilized chicken egg hatches in less than 20 days is 15.87%

b) We want to find out the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch?

[tex]P(X > 22) = 1 - P(X < 22)\\\\P(X > 22) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\P(X > 22) = 1 - P(Z < \frac{22 - 21}{1} )\\\\P(X > 22) = 1 - P(Z < \frac{1}{1} )\\\\P(X > 22) = 1 - P(Z < 1)\\[/tex]

The z-score corresponding to 1 is 0.8413

[tex]P(X > 22) = 1 - 0.8413\\\\P(X > 22) = 0.1587\\\\P(X > 22) = 15.87 \%[/tex]

Therefore, the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch is 15.87%

c) We want to find out the probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days?

[tex]P(19 < X < 21) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(19 < X < 21) = P( \frac{19-21}{1} < Z < \frac{21 - 21}{1} )\\\\P(19 < X < 21) = P( \frac{-2}{1} < Z < \frac{0}{1} )\\\\P(19 < X < 21) = P( -2 < Z < 0 )\\[/tex]

The z-score corresponding to -2 is 0.0227 and 0 is 0.50

[tex]P(19 < X < 21) = P( Z < 0 ) - P( Z < -2 ) \\\\P(19 < X < 21) = 0.50 - 0.0227 \\\\P(19 < X < 21) = 0.4723\\\\P(19 < X < 21) = 47.23 \%[/tex]

Therefore, the probability that a child spends more than 4 hours and less than 8 hours per day unsupervised is 47.23%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.