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Chapter 5: Problem 3

Because serum cholesterol is related to age and sex, some investigators prefer to express it in terms of \(z\) -scores. If \(X=\) raw serum cholesterol, then $$Z=\frac{X-\mu}{\sigma}$$, where \(\mu\) is the mean and \(\sigma\) is the standard deviation of serum cholesterol for a given age-gender group. Suppose \(Z\) is regarded as a standard normal random variable. What is \(\operatorname{Pr}(-1.0

Short Answer

Expert verified
The probability is approximately 0.7745.

Step by step solution

01

Understand the Problem

The problem asks us to find the probability that a standard normal random variable \(Z\) falls between \(-1.0\) and \(1.5\). This probability represents the area under the standard normal distribution curve between these two points.
02

Use the Z-table

Since \(Z\) follows a standard normal distribution, we can use the standard normal distribution table (Z-table) to find the probabilities. We will look up the Z-table values for \(Z = -1.0\) and \(Z = 1.5\).
03

Find Individual Probabilities

From the Z-table, find \(P(Z < 1.5)\) and \(P(Z < -1.0)\). Typically, \(P(Z < 1.5)\) is approximately 0.9332 and \(P(Z < -1.0)\) is approximately 0.1587.
04

Calculate the Probability Between the Two Z-values

To find \(\operatorname{Pr}(-1.0 < Z < 1.5)\), subtract \(P(Z < -1.0)\) from \(P(Z < 1.5)\). Therefore, \(\operatorname{Pr}(-1.0 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.0) = 0.9332 - 0.1587 = 0.7745\).

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Key Concepts

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

Z-score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It tells you how many standard deviations an element is from the mean. To calculate the Z-score of a data point, you use the formula:
  • \[ Z = \frac{X - \mu}{\sigma} \]
Where:
  • \(X\) is the value in question.
  • \(\mu\) represents the mean of the data set.
  • \(\sigma\) signifies the standard deviation.
A Z-score can either be negative or positive. A negative Z-score indicates the data point is below the mean, whereas a positive Z-score indicates it is above the mean. By converting serum cholesterol values into Z-scores, researchers can standardize these values, making them easier to compare and understand across different age and gender groups. This adjust for differences in average cholesterol levels and their variability across different groups.
Probability Distribution
A probability distribution is a statistical function that describes all the possible values and probabilities that a random variable can take within a given range. In the context of the Z-score and standard normal distribution, it involves the bell-shaped curve that is symmetric around a mean of 0 and a standard deviation of 1.
Standard normal distribution is a special case where the mean is 0 and the standard deviation is 1, hence the probabilities are distributed over the curve symmetrically.
  • This distribution allows us to compute the likelihood that a random variable, such as a Z-score, will fall within a certain range. This is crucial when you want to find out the probability of observing a data point within given boundaries, like between -1.0 and 1.5.
Knowing the properties of a probability distribution helps you understand and calculate the chances of various outcomes, based entirely on the mathematical characteristics of the distribution itself.
Z-table
The Z-table is a critical tool used in statistics to determine the probability that a standard normal variable will be less than or equal to a given value. It's essentially a lookup table that provides these cumulative probabilities.
To use the Z-table:
  • Locate the row corresponding to the first two digits of your Z-score.
  • Then, move across to the column that corresponds to the second decimal place of your Z-score.
For example, if your Z-score is 1.5, you'll find the row for 1.5 and then read across to get the value, typically 0.9332, which indicates that there's a 93.32% chance a value is below 1.5 in the standard normal distribution.
The Z-table is invaluable for finding probabilities in the context of the standard normal distribution. It demonstrates how Z-scores translate into percentiles, helping estimate probabilities like \( \operatorname{Pr}(-1.0 < Z < 1.5) \) accurately. Understanding how to read and interpret the Z-table is therefore an essential skill in statistics, especially when analyzing or comparing different data sets by standardization.

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Most popular questions from this chapter

What is the difference between a prevalence rate of breast cancer and an incidence rate of breast cancer?

The change in bone-mineral density of the lumbar spine over a 2-year period among women in the alendronate \(5-\mathrm{mg}\) group was \(+3.5 \%\) (a mean increase), with a standard deviation of \(4.2 \%\). Suppose \(10 \%\) of the women assigned to the alendronate 5 -mg group are actually not taking their pills (noncompliers). If noncompliers are assumed to have a similar response as women in the placebo group, what percentage of women complying with the alendronate 5 -mg treatment would be expected to have a clinically significant decline? (Hint: Use the total-probability rule.)

A study was conducted assessing the effect of observer training on measurement of blood pressure based on NHANES data from \(1999-2000\) (Ostchega, et al., [1] ). A goal was that the difference in recorded blood pressure between the observer and trainer be \(\leq 2 \mathrm{mm}\) Hg in absolute value. It was reported that the mean difference in systolic blood pressure (SBP) between observers and trainers (i.e., mean observer SBP minus trainer SBP) = mean (\Delta) was \(0.189 \mathrm{mm} \mathrm{Hg}\) with \(\mathrm{sd}=2.428 \mathrm{mm} \mathrm{Hg}\). If we assume that the distribution of \(\Delta\) is normally distributed, then what \% of (observer, trainer) pairs have a mean difference of \(\geq 2 \mathrm{mm} \mathrm{Hg}\) in absolute value (i.e., either \(\geq 2 \mathrm{mm} \mathrm{Hg}\) or \(\leq-2 \mathrm{mm} \mathrm{Hg}\) )?

Blood pressure readings are known to be highly variable. Suppose we have mean SBP for one individual over \(n\) visits with \(k\) readings per visit \(\left(\bar{X}_{n, k}\right) .\) The variability of \(\left(\bar{X}_{n, k}\right)\) depends on \(n\) and \(k\) and is given by the formula \(\sigma_{w}^{2}=\sigma_{A}^{2} / n+\) \(\sigma^{2} /(n k),\) where \(\sigma_{A}^{2}=\) between visit variability and \(\sigma^{2}=\) within visit variability. For \(30-\) to 49 -year-old Caucasian females, \(\sigma_{A}^{2}=42.9\) and \(\sigma^{2}=12.8 .\) For one individual, we also assume that \(\bar{X}_{n, k}\) is normally distributed about their true long- term mean \(=\mu\) with variance \(=\sigma_{w}^{2}\). Suppose a woman is measured at two visits with two readings per visit. If her true long-term SBP \(=130 \mathrm{mm} \mathrm{Hg}\) then what is the probability that her observed mean SBP is \(\geq 140 \mathrm{mm}\) Hg? (lgnore any continuity correction.) ( Note : By true mean SBP we mean the average SBP over a large number of visits for that subject.)

Serum cholesterol is an important risk factor for coronary disease. We can show that serum cholesterol is approximately normally distributed, with mean \(=219 \mathrm{mg} / \mathrm{dL}\) and standard deviation \(=50 \mathrm{mg} / \mathrm{dL}\). Some investigators believe that only cholesterol levels over \(250 \mathrm{mg} / \mathrm{dL}\) indicate a high-enough risk for heart disease to warrant treatment. What proportion of the population does this group represent?
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