Questions: The incubation time for a breed of chicks is normally distributed with a mean of 22 days and standard deviation of approximately 3 days. Look at the figure below and answer the following questions. If 1,000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.) (a) in 16 to 28 days chicks (b) in 19 to 25 days chicks (c) in 22 days or fewer chicks (d) in 13 to 31 days chicks

The incubation time for a breed of chicks is normally distributed with a mean of 22 days and standard deviation of approximately 3 days. Look at the figure below and answer the following questions. If 1,000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.)

(a) in 16 to 28 days
chicks
(b) in 19 to 25 days
chicks
(c) in 22 days or fewer
chicks
(d) in 13 to 31 days
chicks
Transcript text: The incubation time for a breed of chicks is normally distributed with a mean of 22 days and standard deviation of approximately 3 days. Look at the figure below and answer the following questions. If 1,000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.) (a) in 16 to 28 days $\qquad$ chicks (b) in 19 to 25 days $\qquad$ chicks (c) in 22 days or fewer $\qquad$ chicks (d) in 13 to 31 days $\qquad$ chicks
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Solution

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Solution Steps

Step 1: Identify the Mean and Standard Deviation

The mean incubation time (\(\mu\)) is 22 days, and the standard deviation (\(\sigma\)) is 3 days.

Step 2: Calculate the Z-Scores for the Given Intervals

For each interval, convert the days into Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \]

(a) Interval: 16 to 28 days
  • For 16 days: \( Z = \frac{16 - 22}{3} = -2 \)
  • For 28 days: \( Z = \frac{28 - 22}{3} = 2 \)
(b) Interval: 19 to 25 days
  • For 19 days: \( Z = \frac{19 - 22}{3} = -1 \)
  • For 25 days: \( Z = \frac{25 - 22}{3} = 1 \)
(c) Interval: 22 days or fewer
  • For 22 days: \( Z = \frac{22 - 22}{3} = 0 \)
Step 3: Determine the Proportion of the Normal Distribution for Each Interval

Using the standard normal distribution table or the given area under the normal curve:

(a) Interval: 16 to 28 days
  • From \( Z = -2 \) to \( Z = 2 \), the area is approximately 95%.
(b) Interval: 19 to 25 days
  • From \( Z = -1 \) to \( Z = 1 \), the area is approximately 68%.
(c) Interval: 22 days or fewer
  • From \( Z = -\infty \) to \( Z = 0 \), the area is 50%.
Step 4: Calculate the Number of Chicks for Each Interval

Given 1,000 eggs:

(a) Interval: 16 to 28 days

\[ 0.95 \times 1000 = 950 \text{ chicks} \]

(b) Interval: 19 to 25 days

\[ 0.68 \times 1000 = 680 \text{ chicks} \]

(c) Interval: 22 days or fewer

\[ 0.50 \times 1000 = 500 \text{ chicks} \]

Final Answer

  • (a) 950 chicks
  • (b) 680 chicks
  • (c) 500 chicks
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