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The length of gestation for swine is normally distributed with mean 114 days and standard deviation 0.75 day. Find the probability that a litter will be born within one day of the mean of \(114 .\)

Short Answer

Expert verified
The probability is 0.8164.

Step by step solution

01

Define the Random Variable

Let the random variable \( X \) represent the gestation length of swine, which follows a normal distribution. That is, \( X \sim N(114, 0.75^2) \), where 114 is the mean and 0.75 is the standard deviation.
02

Identify the Range of Interest

We need to find the probability that the gestation length is within one day of 114. This translates to finding \( P(113 \leq X \leq 115) \).
03

Standardize the Values

Convert the range \(113 \leq X \leq 115\) to the standard normal variable \(Z\) using the formula \( Z = \frac{X - \mu}{\sigma} \). Here, \(\mu = 114\) and \(\sigma = 0.75\). Calculate \( Z_1 = \frac{113 - 114}{0.75}\) and \( Z_2 = \frac{115 - 114}{0.75}\).
04

Calculate the Z-scores

For \( X = 113 \), \( Z_1 = \frac{113 - 114}{0.75} = -\frac{1}{0.75} = -1.33\). For \( X = 115 \), \( Z_2 = \frac{115 - 114}{0.75} = \frac{1}{0.75} = 1.33\).
05

Use Standard Normal Distribution Table

Look up the values of \( P(Z < -1.33) \) and \( P(Z < 1.33) \) in the standard normal distribution table. \( P(Z < 1.33) \approx 0.9082 \) and \( P(Z < -1.33) \approx 0.0918 \).
06

Calculate the Probability

The probability that a litter will be born within one day of the mean of 114 is \( P(113 \leq X \leq 115) = P(Z < 1.33) - P(Z < -1.33) = 0.9082 - 0.0918 = 0.8164 \).

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Key Concepts

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

Standard Deviation
Standard deviation is a key concept in statistics and measures how spread out numbers are in a data set. In simpler terms, it tells us how much the values differ from the mean or average. If the standard deviation is small, most numbers are close to the mean. If it is large, the numbers are more spread out in the data set.

In our exercise about swine gestation, the standard deviation indicates how much the gestation periods of various swine differ from the average gestation of 114 days. In this case, the standard deviation is 0.75 days, suggesting minimal variation in duration between different swine pregnancies.

Knowing the standard deviation helps us make predictions. For instance, with a normal distribution, about 68% of values fall within one standard deviation of the mean, 114 days in this context. This highlights its importance in probability calculation and understanding data variability.
Probability Calculation
Probability calculation in this context involves finding the likelihood of a swine's gestation length falling within a specified range. Probabilities are values ranging from 0 to 1, indicating the chances of an event occurring, where 0 means impossible and 1 means certain.

To find this probability, we first define the range of interest. In the given exercise, we want the gestation length between 113 and 115 days. The process involves converting our normal distribution values into a standard normal form to simplify calculations.

The conversion steps use a formula to turn a normal variable into a z-score, which is easier to work with. This standardization allows us to utilize a standard normal distribution table, speeding up probability calculations. By comparing the z-scores (-1.33 and 1.33 in our example), we can quickly find the probability that a gestation period is within the desired range, which results in a probability of 0.8164 or 81.64%.
Standard Normal Distribution
The standard normal distribution is a special kind of normal distribution where the mean is 0 and the standard deviation is 1. It simplifies the process of finding probabilities associated with any normal distribution. By converting any value from a normal distribution into the standard normal distribution using z-scores, one can easily compute probabilities using standard normal distribution tables.

In the swine gestation problem, we first identified the gestation periods and their deviation from the mean. Then, we calculated the z-scores by standardizing the values of interest. This conversion turns our data into something the standard normal distribution table can process.

The table is powerful. Once our data is in the form of z-scores, the table quickly reveals the probability of data falling below certain z-values. By calculating the difference between the scores, we identified the probability of the gestation period falling within our sought-after range, demonstrating the utility of the standard normal distribution in probability calculations.

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Most popular questions from this chapter

The amount of non-mortgage debt per household for households in a particular income bracket in one part of the country is normally distributed with mean \(\$ 28,350\) and standard deviation \(\$ 3,425 .\) Find the probability that a randomly selected such household has between \(\$ 20,000\) and \(\$ 30,000\) in non-mortgage debt.

The systolic blood pressure \(X\) of adults in a region is normally distributed with mean \(112 \mathrm{~mm} \mathrm{Hg}\) and standard deviation \(15 \mathrm{~mm}\) Hg. A person is considered "prehypertensive" if his systolic blood pressure is between 120 and \(130 \mathrm{~mm} \mathrm{Hg}\). Find the probability that the blood pressure of a randomly selected person is prehypertensive.

The quantity of gasoline purchased in a single sale at a chain of filling stations in a certain region is normally distributed with mean 11.6 gallons and standard deviation 2.78 gallons. Find the three quartiles for the quantity of gasoline purchased in a single sale.

The amount \(X\) of orange juice in a randomly selected half-gallon container varies according to a normal distribution with mean 64 ounces and standard deviation 0.25 ounce. a. Sketch the graph of the density function for \(X\). b. What proportion of all containers contain less than a half gallon (64 ounces)? Explain. c. What is the median amount of orange juice in such containers? Explain.

A machine for filling 2 -liter bottles of soft drink delivers an amount to each bottle that varies from bottle to bottle according to a normal distribution with standard deviation 0.002 liter and mean whatever amount the machine is set to deliver. a. If the machine is set to deliver 2 liters (so the mean amount delivered is 2 liters) what proportion of the bottles will contain at least 2 liters of soft drink? b. Find the minimum setting of the mean amount delivered by the machine so that at least \(99 \%\) of all bottles will contain at least 2 liters.

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