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The mean incubation time of fertilized chicken eggs kept at \(100.5^{\circ} \mathrm{F}\) in a still-air incubator is 21 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day. Source: University of Illinois Extension (a) Draw a normal model that describes egg incubation times of fertilized chicken eggs. (b) Find and interpret the probability that a randomly selected fertilized chicken egg hatches in less than 20 days. (c) Find and interpret the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch. (d) Find and interpret the probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days. (e) Would it be unusual for an egg to hatch in less than 18 days? Why?

Step by step solution

01

Title - Define the Normal Distribution

The incubation times of fertilized chicken eggs form a normal distribution. The mean incubation time is \(\mu = 21\) days and the standard deviation is \(\sigma = 1\) day.

02

Title - Draw the Normal Distribution

Draw a bell curve centered at the mean, 21 days. Mark the standard deviations at 20 days (\(21 - 1\)), 22 days (\(21 + 1\)), 19 days (\(21 - 2\)), and 23 days (\(21 + 2\)). This visualizes the normal model for egg incubation times.

03

Title - Calculate Z-value for less than 20 days (part b)

Calculate the Z-value for 20 days using the formula: \(Z = \frac{X - \mu}{\sigma} = \frac{20 - 21}{1} = -1\). Locate the corresponding probability in the standard normal distribution table: P(Z < -1) ≈ 0.1587.

04

Title - Interpret the Probability for less than 20 days (part b)

The probability that a randomly selected fertilized chicken egg hatches in less than 20 days is approximately 0.1587 or 15.87%. This indicates that around 15.87% of the eggs will hatch in under 20 days.

05

Title - Calculate Z-value for over 22 days (part c)

Calculate the Z-value for 22 days using the formula: \(Z = \frac{X - \mu}{\sigma} = \frac{22 - 21}{1} = 1\). Locate the corresponding probability in the standard normal distribution table: P(Z > 1) ≈ 1 - 0.8413 = 0.1587.

06

Title - Interpret the Probability for over 22 days (part c)

The probability that a randomly selected fertilized chicken egg takes over 22 days to hatch is approximately 0.1587 or 15.87%. This indicates that around 15.87% of the eggs will take over 22 days to hatch.

07

Title - Calculate Probability between 19 and 21 days (part d)

Calculate the Z-value for 19 days: \(Z = \frac{19 - 21}{1} = -2\). Locate the corresponding probability in the standard normal distribution table: P(Z < -2) ≈ 0.0228. The probability for 21 days (mean) is 0.5. Thus, P(19 < Z < 21) = 0.5 - 0.0228 = 0.4772.

08

Title - Interpret Probability between 19 and 21 days (part d)

The probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days is approximately 0.4772 or 47.72%. This indicates that around 47.72% of the eggs will hatch between 19 and 21 days.

09

Title - Calculate Probability for hatching in less than 18 days (part e)

Calculate the Z-value for 18 days: \(Z = \frac{18 - 21}{1} = -3\). Locate the corresponding probability in the standard normal distribution table: P(Z < -3) ≈ 0.0013.

10

Title - Interpret the Unusualness of hatching in less than 18 days (part e)

The probability that a fertilized chicken egg hatches in less than 18 days is approximately 0.0013 or 0.13%. This is a very low probability, indicating that it would be extremely unusual for an egg to hatch in less than 18 days.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mean

In statistics, the mean is the average value in a dataset or distribution. For our chicken egg incubation problem, the mean, denoted by \( \mu \ \), represents the average number of days it takes for an egg to hatch. Here, the mean incubation time is 21 days. This means that, on average, it takes 21 days for a fertilized chicken egg to hatch under the given conditions. The mean is a central value where most of the data points cluster around in a normal distribution.

standard deviation

The standard deviation, represented by \( \sigma \ \), measures the spread or dispersion of a set of data points from the mean. In the context of our exercise, the standard deviation is 1 day. This means that the incubation times of fertilized chicken eggs typically deviate by 1 day from the average incubation time of 21 days. A smaller standard deviation would indicate that the incubation times are closely packed around the mean, while a larger standard deviation would suggest a wider range of incubation times.

probability calculation

Calculating probabilities for a normal distribution involves finding the area under the curve. Here's how you can interpret the probabilities from the given problem:

• Less than 20 days: For this, we first find the Z-value which represents the number of standard deviations away from the mean. The formula used is: \( Z = \frac{X - \mu}{ \sigma } \). Substituting, we get \( Z = \frac{ 20 - 21 }{ 1 } = -1 \). Using a Z-table, the probability for \( Z < -1 \ \) is 0.1587 or 15.87%.
• More than 22 days: Similarly, the Z-value is \( Z = \frac{ 22 - 21 }{ 1 } = 1 \). The probability for \( Z > 1 \) is approximately 0.1587 or 15.87% which we get by \(1 - P(Z < 1) \).
• Between 19 and 21 days: Here, we calculate two Z-values and then find the difference of their cumulative probabilities. For 19 days, \( Z = \frac{ 19 - 21 }{ 1 } = -2 \), corresponding to a probability of 0.0228. For 21 days, the Z-value is 0, corresponding to a probability of 0.5. Thus, the result is \(0.5 - 0.0228 = 0.4772 \) or 47.72%.

Z-value

A Z-value or Z-score represents the number of standard deviations a data point is from the mean. The formula to calculate the Z-value is: \( Z = \frac{X - \mu}{ \sigma } \). It provides a way to standardize different datasets for comparison. In the context of the exercise:

• Less than 20 days: The Z-value is -1, indicating that 20 days is 1 standard deviation below the mean.
• More than 22 days: The Z-value is 1, indicating that 22 days is 1 standard deviation above the mean.
• Between 19 and 21 days: The Z-values are -2 for 19 days and 0 for 21 days, reflecting the spread from the mean for the given incubation periods.

Using Z-values allows you to find probabilities and make inferences about data points relative to the overall distribution.