/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Pregnancy length The length of h... [FREE SOLUTION] | 91Ó°ÊÓ

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Pregnancy length The length of human pregnancies from conception to birth follows a Normal distribution with mean 266 days and standard deviation 16 days. Choose a pregnant woman at random. Let \(X\) \(=\) the length of her pregnancy. (a) Find \(P(X \geq 240)\) and interpret the result. (b) What is \(P(X>240)\) ? Explain. (c) Find the value of \(c\) such that \(P(X \geq c)=0.20 .\) Show your work.

Short Answer

Expert verified
(a) 0.9474; 94.74% of pregnancies are at least 240 days long. (b) 0.9474; same logic. (c) \(c \approx 252.56 \) days.

Step by step solution

01

Identify the parameters and distribution

We are given that the length of human pregnancies follows a Normal distribution with mean \( \mu = 266 \) days and standard deviation \( \sigma = 16 \) days. We denote the random variable representing pregnancy length as \( X \sim N(266, 16^2) \).
02

Standardize the variable

To find probabilities related to \( X \), we standardize to the Z-distribution using the formula: \( Z = \frac{X - \mu}{\sigma} \). For part (a), with \( X = 240 \), we get \( Z = \frac{240 - 266}{16} = -1.625 \).
03

Find \( P(X \geq 240) \)

The probability \( P(X \geq 240) \) is equivalently \( P(Z \geq -1.625) \). Using the Z-table, or a normal distribution calculator, \( P(Z > -1.625) = 0.9474 \). Thus, \( P(X \geq 240) = 0.9474 \). This means 94.74% of pregnancies last at least 240 days.
04

Interpret \( P(X \geq 240) \)

The result indicates that approximately 94.74% of pregnancies are at least 240 days long.
05

Find \( P(X > 240) \)

Since for continuous distributions, \( P(X \geq c) = P(X > c) \), we have \( P(X > 240) = 0.9474 \). This is because the probability of an exact value in a continuous distribution is 0.
06

Set up equation for \( P(X \geq c) = 0.20 \)

We need to find \( c \) such that \( P(X \geq c) = 0.20 \), or equivalently \( P(Z \geq \frac{c-266}{16}) = 0.20 \). This rearranges to \( P(Z less z) = 0.20 \), where \( z = \frac{c-266}{16} \).
07

Find the Z-score corresponding to 0.20 in the tail

From the Z-table, or inverse normal calculation, the Z-score that leaves 0.20 in the upper tail is approximately \( -0.84 \). Thus, \( \frac{c-266}{16} = -0.84 \).
08

Solve for \( c \)

Using the Z-score equation, solve for \( c \):\[\frac{c-266}{16} = -0.84 \Rightarrow c-266 = -0.84 \times 16 \Rightarrow c - 266 = -13.44 \Rightarrow c = 266 - 13.44 = 252.56 \]Thus, \( c \approx 252.56 \).
09

Conclude the value of \( c \)

The value of \( c \) for which \( P(X \geq c) = 0.20 \) is approximately \( 252.56 \) days.

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Key Concepts

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

Z-score
The Z-score is a statistical measure that indicates how many standard deviations a data point is from the mean. It is used to standardize data points within a normal distribution, allowing comparison between different datasets. The formula for calculating the Z-score is:
\[ Z = \frac{X - \mu}{\sigma} \]
where:
  • \(X\) is the data point
  • \(\mu\) is the mean of the dataset
  • \(\sigma\) is the standard deviation
By calculating the Z-score, we transform a data point into a standardized value that represents its position within a normal distribution, which simplifies probability calculations and comparisons.
Probability
Probability is the measure of the likelihood that an event will occur, ranging from 0 to 1. In the context of a normal distribution, it represents the area under the curve for a given range of values. For instance, finding the probability that a pregnancy lasts at least 240 days involves calculating the area from 240 onwards under the normal distribution curve.
This probability helps in understanding the proportion of a dataset falling within specific boundaries. In continuous distributions, the probability of a single exact value is zero since there are infinitely many possible values between any two values in a continuous range.
Standard Deviation
Standard deviation is a measure reflecting the amount of variation or dispersion in a dataset. It indicates how spread out the numbers in a dataset are in relation to the mean. In a normal distribution, most data points are within one standard deviation of the mean. Mathematically, standard deviation is represented as:
\[ \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2} \]
where:
  • \(N\) is the number of observations
  • \(X_i\) is each individual observation
  • \(\mu\) is the mean of the dataset
A smaller standard deviation indicates that data points tend to be closer to the mean, whereas a larger one suggests they are spread over a wider range.
Continuous Distribution
Continuous distributions describe variables that can take on an infinite number of possible values within a given range. Unlike discrete distributions, they deal with measurements rather than counts. A normal distribution is a classic example where the values can be any real number along a continuum.
A key characteristic of continuous distributions is that probabilities are calculated over intervals, since the probability of obtaining any specific value is technically zero. Intervals can be specified in ranges such as \(P(X \geq 240)\), meaning the probability that the variable \(X\) is greater than or equal to 240.
  • These distributions are often represented by a probability density function (PDF) which shows the relative likelihood for a variable.
  • The area under the curve of a PDF for a continuous distribution sums up to 1, representing the total probability space.
Understanding continuous distributions is crucial for working with data that naturally exist on a continuum, such as time, temperature, and in this case, pregnancy length.

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Most popular questions from this chapter

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