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Chapter 2: Problem 10

Compute the following probabilities: u. If \(Y\) is distributed \(N(1.4)\), find \(\operatorname{Pr}(Y \leq 3)\). b. I( \(Y\) is distributed \(N(3,9)\), find \(\operatorname{Pr}(Y>0)\). c. If \(Y\) is distributed \(N(50,25)\), find \(\operatorname{Pr}(40 \leq Y \leq 52)\). d. If \(Y\) is distributed \(N(5,2)\), find \(\operatorname{Pr}(6 \leq Y \leq 8)\).

Short Answer

Expert verified
u: 0.9452, b: 0.8413, c: 0.6326, d: 0.2219.

Step by step solution

01

Understanding Normal Distribution Notation

The notation \(N(\mu, \sigma^2)\) represents a normal distribution with mean \(\mu\) and variance \(\sigma^2\). The standard deviation \(\sigma\) is the square root of the variance.
02

Solving Part u

For \(Y \sim N(1.4)\), this means it's a standard normal distribution since no variance is given, assume variance is 1: \(Y \sim N(1.4, 1)\). We want \(\operatorname{Pr}(Y \leq 3)\). We convert to standard normal: \(Z = \frac{Y - \mu}{\sigma} = \frac{3 - 1.4}{1} = 1.6\). \(\operatorname{Pr}(Y \leq 3) = \operatorname{Pr}(Z \leq 1.6)\). Using standard normal distribution tables or a calculator, \(\operatorname{Pr}(Z \leq 1.6) \approx 0.9452\).
03

Solving Part b

For \(Y \sim N(3,9)\), \(\operatorname{Pr}(Y>0)\) needs to be found. Convert to standard normal: \(Z = \frac{Y - \mu}{\sigma} = \frac{0 - 3}{3} = -1\). \(\operatorname{Pr}(Y > 0) = \operatorname{Pr}(Z > -1) = 1 - \operatorname{Pr}(Z \leq -1)\). From tables, \(\operatorname{Pr}(Z \leq -1) \approx 0.1587\), so \(\operatorname{Pr}(Y > 0) = 0.8413\).
04

Solving Part c

For \(Y \sim N(50,25)\), find \(\operatorname{Pr}(40 \leq Y \leq 52)\). Convert boundaries to standard normal: \(Z_1 = \frac{40 - 50}{5} = -2\) and \(Z_2 = \frac{52 - 50}{5} = 0.4\). \(\operatorname{Pr}(-2 \leq Z \leq 0.4) = \operatorname{Pr}(Z \leq 0.4) - \operatorname{Pr}(Z \leq -2)\). From tables, \(\operatorname{Pr}(Z \leq 0.4) \approx 0.6554\) and \(\operatorname{Pr}(Z \leq -2) \approx 0.0228\). So, \(\operatorname{Pr}(40 \leq Y \leq 52) \approx 0.6326\).
05

Solving Part d

For \(Y \sim N(5,2)\), find \(\operatorname{Pr}(6 \leq Y \leq 8)\). Convert to standard normal: \(Z_1 = \frac{6 - 5}{\sqrt{2}} \approx 0.71\) and \(Z_2 = \frac{8 - 5}{\sqrt{2}} \approx 2.12\). \(\operatorname{Pr}(0.71 \leq Z \leq 2.12) = \operatorname{Pr}(Z \leq 2.12) - \operatorname{Pr}(Z \leq 0.71)\). From tables, \(\operatorname{Pr}(Z \leq 2.12) \approx 0.9830\) and \(\operatorname{Pr}(Z \leq 0.71) \approx 0.7611\). So, \(\operatorname{Pr}(6 \leq Y \leq 8) \approx 0.2219\).

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Key Concepts

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

Probability Calculation
Probability is a measure of the likelihood that a particular event will occur. In the context of a normal distribution, probability helps us understand how likely it is that a particular value or range of values will be observed. When we have a random variable, such as \(Y\), that follows a normal distribution \(N(\mu, \sigma^2)\), we can calculate the probability of \(Y\) falling within a certain interval or above or below a particular point.

To calculate probabilities related to a normal distribution, we often convert our target range (like \(Y \leq 3\)) into its corresponding value on the standard normal distribution using a z-score. This z-score allows us to utilize standard normal distribution tables or calculators effectively.
  • First, determine the mean and variance of the distribution.
  • Convert the value of \(Y\) into a z-score.
  • Use tables or tools to find the probability associated with that z-score.
With the calculated probabilities, we get insights such as the probability of a random variable being less than, greater than, or between certain values.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution. It has a mean \( \mu = 0 \) and a standard deviation \( \sigma = 1 \). The standard normal distribution is used as a reference to simplify probability calculations for any normal distribution, by transforming other normal variables into this form.

Why use it? Because it provides a universal scale to assess probabilities. By converting measures from any normal distribution into z-scores, we use a common method to read probabilities from standard tables or calculate them efficiently.
  • Any value from a normal distribution can be transformed into the standard normal distribution using the formula:
\[ Z = \frac{Y - \mu}{\sigma} \]
  • \( Z \) here represents how many standard deviations \(Y\) is from the mean \( \mu \).
This powerful tool can effortlessly connect different instances of the normal distribution to a single reference model.
Z-Score
A z-score, sometimes referred to as a standard score, indicates how many standard deviations a particular data point is from the mean. It helps us standardize different data points so that we can easily compare them and calculate probabilities using the standard normal distribution.

To compute a z-score:
  • Subtract the mean \( \mu \) of the distribution from the data point of interest \( Y \).
  • Divide the result by the standard deviation \( \sigma \) of the distribution.
The formula looks like this:
\[ Z = \frac{Y - \mu}{\sigma} \]
This simple calculation brings data from any normal distribution into the realm of the standard normal distribution, where the mean is 0, and the standard deviation is 1.

Interpreting z-scores:
  • A z-score of 0 indicates the value is precisely at the mean.
  • Positive z-scores tell us the value is above the mean.
  • Negative z-scores reveal the value is below the mean.
Understanding z-scores enables us to address questions of probability consistently across various datasets and contexts.

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Most popular questions from this chapter

\(X\) is a random variable with moments \(E(X), E\left(X^{2}\right) . E\left(X^{3}\right)\), and so forth. a. Show \(E(X-\mu)^{3}=E\left(X^{3}\right)-3\left[E\left(X^{2}\right)\right][E(X)]+2[E(X)]^{3}\). b. Show \(\left.E(X-\mu)^{4}=E\left(X^{4}\right)-4[E(X)]\left[E\left(X^{3}\right)\right]+6 \mid E(X)\right]^{2}\left[E\left(X^{2}\right)\right]-\) \(3[E(X)]^{4} .\)

This exercise provides an example of a pair of random variables \(X\) and \(Y\) for which the conditional mean of \(Y\) given \(X\) depends on \(X\) but \(\operatorname{corr}(X, Y)=0\). Let \(X\) and \(Z\) be two independently distributed standard normal random variables, and let \(Y=X^{2}+Z\). a. Show that \(E(Y \mid X)=X^{2}\). b. Show that \(\mu_{Y}=1\). c. Show that \(E(X Y)=0\). (Hint: Use the fact that the odd moments of a standard normal random variable are all zero.) d. Show that \(\operatorname{cov}(X, Y)=0\) and thus \(\operatorname{corr}(X, Y)=0\).

In a given population of two-earner male/[cmale couples, male carnings have a mean of \(\$ 40.000\) per year and a standard deviation of \(\$ 12,1000\). Female carnings have a mean of \(\$ 45.000\) per year and a standard deviation of \(\$ 18.000\). The corrclation between male and female earnings for a couple is \(0.80\). Let C denote the combined earnings for a randomly selected couple. a. What is the inean of C? b. What is the covariance between malc and female tarnings? c. What is the standard deviation of \(C\) ? d. Convert the answers to (a)-(c) frons \(\$$ (dollars) to \)\in$ (euros).

\(Y_{i}, i=1, \ldots, n\), arc i.i.d. Bernoulli random variables with \(p=0.4\). Let \(\bar{Y}\) denote the sample mean. a. Use the central limit to compute approximations Ior i. \(\operatorname{Pr}(\bar{Y} \geq 0.43)\) when \(n=100\). ii. \(\operatorname{Pr}(\bar{Y} \leq 0.37)\) when \(n=400\). b. How large would \(n\) need to be to ensure that \(\operatorname{Pr}(0.39 \leq \bar{Y} \leq 0.41) \geq\) \(0.95 ?\) (Use the central limit theorem to compute an approximate answer.)

In any year, the weather can inflict storm damage to a home. From year to year. the damage is random. Let \(Y\) denote the dollar value of damage in any given year. Suppose that in \(95 \%\) of the years \(Y=\$ 0\), but in \(5 \%\) of the years \(Y=\$ 20.000\). a. What is the mean and standard deviation of the damage in any year? b. Consider an "insurance pool" of 100 people whose homes are suffciently dispersed so that, in any year, the damage to different homes can be viewed as independently distributed random variables. Let \(\bar{Y}\) denote the average damage to these 100 bomes in a year. (i) What is the expected value of the average damage \(\bar{Y}\) ? (ii) What is the probability that \(\bar{Y}\) cxceeds \(\$ 2000\) ?
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