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A normal distribution has mean \(\mu=110\) points and standard deviation \(\sigma=12\) points. Find the \(z\) -value of each of the following: (a) \(x=98\) points. (b) \(x=110\) points. (c) \(x=128\) points. (d) \(x=71\) points.

Short Answer

Expert verified
The z-scores for x = 98, 110, 128, and 71 are -1, 0, 1.5, and -3.25 respectively.

Step by step solution

01

Identify given values

From the problem, we know that the mean (μ) is 110 points and the standard deviation (σ) is 12 points.
02

Apply the z-score formula for x = 98

For the first x-value (x = 98), substitute into the z-score formula: z = (x - μ) / σ => z = (98 - 110) / 12. Compute this to find the corresponding z-score.
03

Apply the z-score formula for x = 110

For the second x-value (x = 110), again substitute into the z-score formula: z = (x - μ) / σ => z = (110 - 110) / 12. Compute this to find the corresponding z-score.
04

Apply the z-score formula for x = 128

For the third x-value (x = 128), substitute into the z-score formula: z = (x - μ) / σ => z = (128 - 110) / 12. Compute this to find the corresponding z-score.
05

Apply the z-score formula for x = 71

Finally, for the fourth x-value (x = 71), again substitute into the z-score formula: z = (x - μ) / σ => z = (71 - 110) / 12. Compute this to find its z-score.

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Key Concepts

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

Understanding Z-Score Calculation
The z-score is a statistical measurement that tells you how many standard deviations an element, \( x \), is from the mean \( \mu \) of a normal distribution. When you compute a z-score, you are standardizing the data allowing for comparison across different normal distributions. To find a z-score, you use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Let's explore how to use this formula in practice. If you have a set of data points and you want to know how they relate to the average (mean), calculate each point's z-score. For example, if \( x = 98 \), \( \mu = 110 \), and \( \sigma = 12 \), substitute these values into the formula:
  • \( z = \frac{98 - 110}{12} = \frac{-12}{12} = -1 \)
This tells you that 98 is one standard deviation below the mean of the sample, providing insight into how typical or atypical the value is within that distribution.
Deciphering Standard Deviation
Standard deviation, denoted as \( \sigma \), measures the amount of variability or dispersion in a set of values. A low standard deviation means data points are generally close to the mean, while a high standard deviation indicates they are spread out over a wider range of values.

For instance, in the given problem, a standard deviation of 12 points implies the scores tend to deviate from the mean by about 12 points. Understanding this concept can help contextualize z-scores and gauge how likely a particular data point is in the context of the whole dataset.
  • A small \( \sigma \) suggests a tightly clustered dataset.
  • A large \( \sigma \) indicates more scattered data.
This measure is crucial as it adjusts the scale of z-scores and determines how "rare" or "common" a data point might be within the distribution.
Grasping the Mean in Normal Distribution
The mean, represented as \( \mu \), is the average of all data points in a dataset. In a normal distribution, the mean is not only the measure of central tendency but also the symmetrical center of the distribution curve.

Calculating the mean involves summing all the values in the dataset and then dividing by the number of values. In our exercise scenario, the mean is given as 110 points. This value represents the central point around which the data is distributed.
  • It serves as the reference value for calculating z-scores.
  • It acts as a benchmark for identifying how typical a particular score is.
Recognizing the mean's role helps you understand how individual scores relate to the overall sample context, whether they fall below, above, or right at the average.

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Most popular questions from this chapter

The distribution of weights for six-month-old baby boys has mean \(\mu=8.16 \mathrm{~kg}\) and standard deviation \(\sigma=0.95 \mathrm{~kg}\) (see Exercise 53 ). (a) Suppose that a six-month-old baby boy weighs in the 95th percentile of his age group. Estimate his weight in kilograms rounded to two decimal places. (b) Suppose that a six-month-old baby boy weighs in the 20th percentile of his age group. Estimate his weight in kilograms rounded to two decimal places.

Find the \(z\) -value of (a) the first quartile of a normal distribution. (b) the third quartile of a normal distribution.

Suppose that a random sample of \(n=7056\) adults is to be chosen for a survey. Assume that the gender of each adult in the sample is equally likely to be male as it is female. Estimate the probability that (a) the number of females in the sample is between 3486 and 3570 . (b) the number of females in the sample is less than 3486 . (c) the percentage of females in the sample is below \(50.6 \%\). (Hint: Find \(50.6 \%\) of 7056 first.)

The distribution of weights for 12 -month-old baby boys has mean \(\mu=10.5 \mathrm{~kg}\) and standard deviation \(\sigma=1.2 \mathrm{~kg} .\) (a) Suppose that a 12 -month-old boy weighs \(11.3 \mathrm{~kg}\). Approximately what weight percentile is he in? (b) Suppose that a 12 -month-old boy weighs \(8.1 \mathrm{~kg}\). Approximately what weight percentile is he in? (c) Suppose that a 12 -month-old boy is in the 84 th percentile in weight. Estimate his weight.

A normal distribution has mean \(\mu=30 \mathrm{~kg}\) and standard deviation \(\sigma=15\) kg. Find the z-value of each of the following: (a) \(x=45 \mathrm{~kg}\) (b) \(x=0 \mathrm{~kg}\) (c) \(x=54 \mathrm{~kg}\) (d) \(x=3 \mathrm{~kg}\)

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