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Chapter 17: Problem 15

X has a normal distribution such that the probability that \(x\) is larger than 14.6 is \(93.5 \%\) and \(P(x>29.6)=2.2 \%\). Find the mean \(\mu\) and the standard deviation \(\sigma\) of this distribution.

Short Answer

Expert verified
The mean \( \mu \) is approximately 21.05 and the standard deviation \( \sigma \) is approximately 4.27.

Step by step solution

01

Understand the Problem

We need to find the mean \( \mu \) and the standard deviation \( \sigma \) of a normal distribution where the probability \( P(X > 14.6) = 93.5\% \) and \( P(X > 29.6) = 2.2\% \).
02

Convert Probabilities to Z-scores

For a normal distribution, the Z-score corresponding to a probability \( p \) is found using the Z-table or a calculator. \( P(X > 14.6) = 93.5\% \) corresponds to the Z-score \( z_1 \) such that \( P(Z < z_1) = 1 - 0.935 = 0.065 \). Similarly, \( P(X > 29.6) = 2.2\% \) corresponds to \( z_2 \) such that \( P(Z < z_2) = 1 - 0.022 = 0.978 \).
03

Find Z-scores

Using the Z-table or calculator, find \( z_1 \) for 0.065 which is approximately \( z_1 = -1.51 \), and \( z_2 \) for 0.978 which is approximately \( z_2 = 2.00 \).
04

Set Up the Equations

For these Z-scores, apply the equation of Z, which is \( Z = \frac{X - \mu}{\sigma} \). For \( x = 14.6 \), \( -1.51 = \frac{14.6 - \mu}{\sigma} \) and for \( x = 29.6 \), \( 2.00 = \frac{29.6 - \mu}{\sigma} \).
05

Solve for Mean and Standard Deviation

From the equations, we have two equations with two unknowns: \[\frac{14.6 - \mu}{\sigma} = -1.51 \ \frac{29.6 - \mu}{\sigma} = 2.00 \]Subtract them to eliminate \( \mu \): \[ 2.00 - (-1.51) = \frac{29.6 - 14.6}{\sigma} \] Solve for \( \sigma \): \[ 3.51 = \frac{15}{\sigma} \] \( \sigma = \frac{15}{3.51} \approx 4.27 \). Now, use one equation to solve for \( \mu \): \[ 14.6 - \mu = -1.51 \times 4.27 \]\( \mu = 14.6 + 6.45 \approx 21.05 \).
06

Verify the Solution

Substitute \( \mu = 21.05 \) and \( \sigma = 4.27 \) back into both equations to check for consistency. The calculations should lead you to the original Z-scores of \(-1.51\) and \(2.00\), confirming the solution.

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Key Concepts

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

Z-score
Z-scores are an incredibly useful statistical tool when dealing with normal distributions. They're a way of measuring how many standard deviations a data point is from the mean. Suppose you have a normal distribution with a mean (\( \mu \)) and a standard deviation (\( \sigma \)). A Z-score is a normalized score and is often used to find the probability of a data point falling above or below a specific value.
  • Formula: \( Z = \frac{X - \mu}{\sigma} \)
  • Interpretation: A Z-score of 0 indicates a value that is exactly at the mean. Positive Z-scores indicate values above the mean, and negative Z-scores represent values below the mean.
Z-scores allow for the comparison of data points from different normal distributions. They're also vital in converting probabilities, which can then be applied in probability calculations and statistical analyses. In this exercise, Z-scores help translate percentage probabilities into values we can work with using standard normal distribution tables.
Mean and Standard Deviation
The mean (\( \mu \)) and standard deviation (\( \sigma \)) are parameters that define a normal distribution. They help us understand and interpret statistical data, describing the central tendency (mean) and the variation within the data set (standard deviation).
  • Mean (\( \mu \)): The average or central value of a data set. All data points are summed and then divided by the number of points.
  • Standard Deviation (\( \sigma \)): Measures the amount of variation or dispersion in a set of values. A lower value means data points are closer to the mean, while a higher value indicates a wider spread.
In normal distributions, understanding the mean and standard deviation is crucial, as they directly influence the shape of the distribution.
For the normal distribution problem above, calculating these values allows us to further perform probability calculations and understand data dynamics.
Probability Calculations
Probability calculations in the context of normal distributions often involve using Z-scores to find probabilities associated with specific data points. Here's a simple guide:
  • Identify the value of \( X \) you're interested in.
  • Convert \( X \) into a Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \).
  • Use a Z-table, calculator, or statistical software to find the probability associated with that Z-score.
This process helps calculate the probability that a variable falls within or outside a range of values.
In the given exercise, we predicted probabilities that reveal how data points compare to others within the distribution. Converting probabilities and working backward using Z-scores helped us understand the mean and standard deviation of the distribution.

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Most popular questions from this chapter

A closed box contains eight red balls and four white ones. A ball is taken out at random, its colour noted, and then returned. This is done three times. Let \(x\) represent the number of red balls drawn. a) Set up a table to show the probability distribution of \(X\). b) What is the expected number of red balls in this experiment?

A supplier of copper wire looks for flaws before despatching it to customers. It is known that the number of flaws follows a Poisson probability distribution with a mean of 2.3 flaws per metre. a) Determine the probability that there are exactly 2 flaws in 1 metre of the wire. b) Determine the probability that there is at least one flaw in 2 metres of the wire.

The scores on a public schools examination are normally distributed with a mean of 550 and a standard deviation of 100 a) What is the probability that a randomly chosen student from this population scores below 400? b) What is the probability that a student will score between 450 and \(650 ?\) c) What score should you have in order to be in the 90 th percentile? d) Find the IQR of this distribution.

X has a normal distribution with mean \(16 .\) Given that the probability that \(x\) is less than 16.56 is \(64 \%,\) find the standard deviation, \(\sigma,\) of this distribution.

\(X\) has probability distribution as shown in the table. $$\begin{array}{|l|c|c|c|c|c|} \hline x & 5 & 10 & 15 & 20 & 25 \\ \hline P(x) & \frac{3}{20} & \frac{7}{30} & k & \frac{3}{10} & \frac{13}{60} \\\ \hline \end{array}$$ a) Find the value of \(k\) b) Find \(P(x>10)\) c) Find \(P(5
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