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Chapter 17: Problem 16

\(X \sim N\left(\mu, \sigma^{2}\right) . P(x>19.6)=0.16\) and \(P(x<17.6)=0.012 .\) Find \(\mu\) and \(\sigma\)

Short Answer

Expert verified
The mean is approximately 18.99 and the standard deviation is 0.615.

Step by step solution

01

Understand the Given Information

You are given a normal distribution with random variable \(X\) which follows a normal distribution \(N(\mu, \sigma^2)\) where \(\mu\) is the mean and \(\sigma^2\) is the variance. The problem also provides \(P(x>19.6)=0.16\) and \(P(x<17.6)=0.012\). We must find \(\mu\) and \(\sigma\).
02

Convert Probabilities to Z-Scores

Utilize the standard normal distribution tables to find the Z-scores corresponding to the given probabilities. For \(P(x>19.6)=0.16\), it means 19.6 is the 84th percentile since \(1-0.16 = 0.84\). Consult a Z-table to find the Z-score for 0.84, which is approximately 0.995. Similarly, for \(P(x<17.6)=0.012\), the Z-score corresponding to 0.012 is approximately -2.260.
03

Relate Z-Scores to \(X\)

Use the formula for Z-scores: \(Z = \frac{x - \mu}{\sigma}\). For \(x = 19.6\), \(0.995 = \frac{19.6 - \mu}{\sigma}\) (Equation 1). For \(x = 17.6\), \(-2.260 = \frac{17.6 - \mu}{\sigma}\) (Equation 2).
04

Solve the System of Equations

Solve the two equations simultaneously. From Equation 1, \(19.6 - \mu = 0.995 \sigma\). From Equation 2, \(17.6 - \mu = -2.260 \sigma\). Subtract these two equations: \((19.6 - \mu) - (17.6 - \mu) = 0.995 \sigma + 2.260 \sigma\). Simplifying gives \(2 = 3.255 \sigma\), so \(\sigma = \frac{2}{3.255} \approx 0.615.\)
05

Find \(\mu\) Using \(\sigma\)

Substitute \(\sigma\) back into either equation to find \(\mu\). Using Equation 1, \(19.6 - \mu = 0.995 \times 0.615\) which simplifies to \(19.6 - \mu = 0.6129\). Solving for \(\mu\) gives \(\mu = 19.6 - 0.6129 = 18.9871\).
06

Verify the Solution

Use the calculated \(\mu = 18.9871\) and \(\sigma = 0.615\) in the original probabilities to ensure satisfaction. Compute \(Z\) for 19.6 and 17.6 to check if they approximately match the Z-scores of 0.995 and -2.260, validating the solution.

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Key Concepts

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

Z-scores
Z-scores are a way to standardize a data point within a distribution, converting it into a measure of standard deviations away from the mean of the distribution. This conversion is particularly useful in a normal distribution setting.

To calculate the Z-score for a data point, use the formula:
  • \(Z = \frac{x - \mu}{\sigma}\)
Here, \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

In our context, the Z-scores were necessary to translate the probability statements given in the exercise into specific values that can be used to find the unknown mean and standard deviation. The conversion to Z-scores allows us to use standard normal distribution tables, which are based on probabilities and corresponding Z-scores, to find these unknown values.
Mean and Variance
The mean, represented by \(\mu\), is a measure of central tendency, indicating the average of all data points in a distribution. In a normal distribution, the mean is located at the center of the bell-shaped curve.

The variance, expressed as \(\sigma^2\), measures the spread of the data points around the mean. The standard deviation, \(\sigma\), is simply the square root of the variance and provides a more intuitive measure of spread. It reflects how much individual data points differ from the mean.

Finding \(\mu\) and \(\sigma\) in this problem involves understanding how these parameters relate to the probabilities provided. With variance linking directly to standard deviation, solving for these parameters becomes a matter of translating probability conditions into equations that describe the mean and variance vis-a-vis given data points.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution. It has a mean \(\mu\) of 0 and a standard deviation \(\sigma\) of 1. This distribution is significant in statistical analyses because it allows researchers to convert complex data into a manageable format using Z-scores.

Converting any given normal distribution into a standard normal distribution involves two key processes: subtracting the mean from any data point and then dividing by the standard deviation. This process results in a Z-score that fits within the framework of the standard normal distribution.

In exercises like the one presented, the standard normal distribution is accessed via tables or technology, enabling the solutions for probabilities and percentiles for any normal distribution to be derived systematically. These resources simplify the problem-solving by standardizing the distribution, which was vital in calculating \(\mu\) and \(\sigma\) from the probabilities given.

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Most popular questions from this chapter

The number of faults in the knit of a certain fabric has an average of 4.4 faults per square metre. It is also assumed to have a Poisson distribution. a) Find the probability that a \(1 \mathrm{m}^{2}\) piece of this fabric contains at least 1 fault. b) Find the probability that a \(3 \mathrm{m}^{2}\) piece of this fabric contains at least 1 fault. c) Find the probability that three \(1 \mathrm{m}^{2}\) pieces of this fabric contain 1 fault.

Dogs have health insurance too! Owners of dogs in many countries buy health insurance for their dogs. \(3 \%\) of all dogs have health insurance. In a random sample of 100 dogs in a large city, find a) the expected number of dogs with health insurance b) the probability that 5 of the dogs have health insurance c) the probability that more than 10 dogs have health insurance.

X has a normal distribution with mean \(91 .\) Given that the probability that \(x\) is larger than 104 is \(24.6 \%,\) find the standard deviation \(\sigma\) of this distribution.

Chicken eggs are graded by size for the purpose of sales. In Europe, modern egg sizes are defined as follows: very large has a mass of \(73 \mathrm{g}\) or more, large is between 63 and \(73 \mathrm{g},\) medium is between 53 and \(63 \mathrm{g},\) and small is less than 53 g. The small size is usually considered as undesirable by consumers. a) Mature hens (older than 1 year) produce eggs with an average mass of \(67 \mathrm{g} .98 \%\) of the eggs produced by mature hens are above the minimum desirable weight. What is the standard deviation if the egg production can be modelled by a normal distribution? b) Young hens produce eggs with a mean mass of \(51 \mathrm{g}\). Only \(28 \%\) of their eggs exceed the desired minimum. What is the standard deviation? c) A farmer finds that \(7 \%\) of his farm's eggs are 'underweight; and \(12 \%\) are very large. Estimate the mean and standard deviation of this farmer's eggs.

A machine is set to produce bags of salt, whose weights are distributed normally, with a mean of \(110 \mathrm{g}\) and standard deviation of \(1.142 \mathrm{g}\). If the weight of a bag of salt is less than \(108 \mathrm{g}\), the bag is rejected. With these settings, \(4 \%\) of the bags are rejected. The settings of the machine are altered and it is found that \(7 \%\) of the bags are rejected. a) (i) If the mean has not changed, find the new standard deviation, correct to three decimal places. The machine is adjusted to operate with this new value of the standard deviation. (ii) Find the value, correct to two decimal places, at which the mean should be set so that only \(4 \%\) of the bags are rejected. b) With the new settings from part a), it is found that \(80 \%\) of the bags of salt have a weight which lies between \(A \mathrm{g}\) and \(\mathrm{Bg}\), where \(A\) and \(B\) are symmetric about the mean. Find the values of \(A\) and \(B\), giving your answers correct to two decimal places.
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