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Chapter 5: Problem 57

In a normal distribution with \(\mu=-15\) and \(\sigma=0.4\) find the \(x\) -value that corresponds to the a) 46 th percentile b) 92 nd percentile

Short Answer

Expert verified
46th percentile: x = -15.04 92nd percentile: x = -14.436

Step by step solution

01

Identify Percentile for Part (a)

The 46th percentile corresponds to a cumulative probability of 0.46 in a standard normal distribution, because percentiles represent the probability of a value being below a certain point.
02

Find z-score for Part (a)

Using a standard normal distribution table or a calculator with an inverse normal distribution function, find the z-score that corresponds to a cumulative probability of 0.46. This approximately gives us a z-score of -0.10.
03

Transform z-score to x-value for Part (a)

Using the formula for transforming a z-score to an x-value in a normal distribution, which is given by: \[ x = ext{mean} + z imes ext{standard deviation} \]For this problem, \( x = -15 + (-0.10) imes 0.4 \)Calculate to find \( x \).
04

Calculate Transformation for Part (a)

Substitute the values into the transformation formula:\[ x = -15 + (-0.10) imes 0.4 = -15 - 0.04 = -15.04 \]Thus, the x-value for the 46th percentile is -15.04.
05

Identify Percentile for Part (b)

The 92nd percentile means the cumulative probability is 0.92. We need to find the corresponding z-score for this cumulative probability.
06

Find z-score for Part (b)

Look up the z-score corresponding to 0.92 in the standard normal distribution table or using an inverse norm function in a calculator. This gives us approximately a z-score of 1.41.
07

Transform z-score to x-value for Part (b)

Use the same transformation formula: \[ x = ext{mean} + z imes ext{standard deviation} \]For the 92nd percentile, substitute the known values:\( x = -15 + 1.41 imes 0.4 \)Calculate the x-value.
08

Calculate Transformation for Part (b)

Perform the calculation:\[ x = -15 + 1.41 imes 0.4 = -15 + 0.564 = -14.436 \]So, the x-value for the 92nd percentile is -14.436.

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Key Concepts

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

Percentile Calculation
Understanding how percentiles work is important when dealing with normal distributions. A percentile shows the value below which a given percentage of observations fall in a dataset. For instance, the 46th percentile is the value below which 46% of the data lie.
In a normal distribution, percentiles are equated with cumulative probabilities. On the normal curve, the area to the left of a value represents its percentile as a percentage.
To calculate the percentile in a normal distribution:
  • Determine the cumulative probability for the required percentile. For example, a 46th percentile has a cumulative probability of 0.46.
  • Identify the z-score corresponding to this cumulative probability using a standard normal distribution table or calculator.
  • Finally, convert that z-score to an actual data value (x-value) using the mean and standard deviation of your distribution.
This process allows you to find the exact data point associated with any given percentile in a normal distribution.
Z-Score Transformation
Z-score transformation is a method of translating a value from a normal distribution into the standard normal distribution, which has a mean of 0 and a standard deviation of 1. This transformation is useful for identifying where an observation lies in a dataset.
The transformation is governed by the formula:
  • Calculate the z-score by using the formula \[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation of the original dataset.
  • To find a specific data value from a z-score, use the reverse transformation: \[ x = \mu + z \times \sigma \]
This reverse operation allows us to convert a z-score back to a value in the original dataset. Understanding z-score transformations are key in comparing values from different normal distributions.
Cumulative Probability
Cumulative probability in a normal distribution represents the likelihood that a variable takes a value less than or equal to a certain point. For each value, cumulative probability is depicted by the area under the normal curve to the left of that value.
To comprehend cumulative probabilities:
  • Cumulative probabilities are utilized to find percentiles. For a value at a cumulative probability of 0.92, it lies at the 92nd percentile, meaning 92% of the data fall below it.
  • Using the standard normal distribution table or software, you can find cumulative probabilities for various z-scores. These values help in determining where a specific value ranks within the distribution.
  • Once you have the cumulative probability, you can find the corresponding z-score and transform it to get the actual data point in your distribution.
Understanding cumulative probability is essential for making accurate predictions and interpretations based on the normal distribution model.

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Most popular questions from this chapter

A dart is thrown at a number line in such a way that it always lands in [0,10] . Let \(x\) represent the number the dart hits. Suppose the probability density function for \(x\) is given by \(f(x)=\frac{1}{50} x, \quad\) for \(0 \leq x \leq 10\) Find \(P(2 \leq x \leq 6),\) the probability that the dart lands in [2,6]

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is inversely proportional to the square root of \(y\).

For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{1}{5}, \quad[3,8] $$

The capitalized cost, \(c,\) of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed with the formula $$ c=c_{0}+\int_{0}^{L} m(t) e^{-k t} d t $$ where \(c_{0}\) is the initial cost of the asset, \(L\) is the lifetime (in years), \(k\) is the interest rate (compounded continuously), and \(m(t)\) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. $$ \begin{array}{l} c_{0}=\$ 300,000, k=5 \%, m(t)=\$ 30,000+\$ 500 t, \\ L=20 \end{array} $$

Test score distribution. In a large class, student test scores had a mean of \(\mu=76\) and a standard deviation \(\sigma=7\) a) The top \(12 \%\) of students got an \(A\). Find the minimum score needed to get an \(\mathrm{A}\) (round to the appropriate integer). b) The top \(75 \%\) of students passed. Find the minimum score needed to pass (round to the appropriate integer).
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