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There is no nice formula for the standard normal cdf \(\Phi(z)\), but several good approximations have been published in articles. The following is from "Approximations for Hand Calculators Using Small Integer Coefficients" (Math. Comput., 1977: \(214-222\) ). For \(05)\)

Short Answer

Expert verified
a. 0.142; b. 0.9988; c. 0.9999836; d. 6.39 × 10^{-9}.

Step by step solution

01

Understand the Problem

The task is to use the given approximation formula for the standard normal probability to calculate specific probabilities. We are to compare these with exact values if possible using standard normal distribution tables.
02

Calculate P(Z ≥ 1) using the formula

For part (a), we want to find the probability that a standard normal variable is greater than or equal to 1. Using the approximation:\[ P(Z \geq z) \approx 0.5 \exp \left\{-\left[\frac{(83 \times 1+351) \times 1+562}{(703 / 1)+165}\right]\right\} \]Performing the calculations step by step:- Numerator: \((83 \times 1 + 351) \times 1 + 562 = 830 + 351 + 562 = 943.5\)- Denominator: \(703 + 165 = 868\)- Exponent: \(-\left[ \frac{943.5}{868} \right] \approx -1.086\)- Finally: \(0.5 \times \exp(-1.086) \approx 0.142\)
03

Explain that P(Z < -3) is equivalent to 1 - P(Z ≥ 3)

For part (b), the probability that Z is less than -3 is the same as 1 minus the probability that Z is greater than 3, since standard normal distribution is symmetric around 0.
04

Calculate P(Z ≥ 3) using the formula

Using the same steps as before but for \(z = 3\):\[ P(Z \geq 3) \approx 0.5 \exp \left\{-\left[\frac{(83 \times 3 + 351) \times 3 + 562}{(703 / 3)+165}\right]\right\} \]- Numerator: \((83 \times 3 + 351) \times 3 + 562 = 784.5 + 351 + 562 = 2692.5\)- Denominator: \(703 / 3 + 165 = 238.67 + 165 = 403.67\)- Exponent: \(-\left[ \frac{2692.5}{403.67} \right] \approx -6.67\)- Finally: \(0.5 \times \exp(-6.67) \approx 0.0012\)Thus, \(P(Z < -3) = 1 - 0.0012 = 0.9988\).
05

Calculate P(-4 < Z < 4) using symmetry and complementary probabilities

For part (c), use both tails to find \(P(-4 < Z < 4)\):\[ P(-4 < Z < 4) = 1 - P(|Z| \geq 4) = 1 - [P(Z \geq 4) + P(Z \leq -4)] \]Since the distribution is symmetric, \(P(Z \geq 4) = P(Z \leq -4)\).Calculate \(P(Z \geq 4)\):\[ P(Z \geq 4) \approx 0.5 \exp \left\{-\left[\frac{(83 \times 4 + 351) \times 4 + 562}{(703 / 4)+165}\right]\right\} \]- Numerator: \((83 \times 4 + 351) \times 4 + 562 = 3845.5\)- Denominator: \(175.75 + 165 = 340.75\)- Exponent: \(-\left[ \frac{3845.5}{340.75} \right] \approx -11.28\)- Finally: \(0.5 \times \exp(-11.28) \approx 0.00000821\)Therefore, \(P(-4 < Z < 4) \approx 1 - 2\times 0.00000821 \approx 0.9999836\).
06

Calculate P(Z > 5) using the formula

For part (d), directly calculate \(P(Z > 5)\):\[ P(Z \geq 5) \approx 0.5 \exp \left\{-\left[\frac{(83 \times 5 + 351) \times 5 + 562}{(703 / 5)+165}\right]\right\} \] - Numerator: \((83 \times 5 + 351) \times 5 + 562 = 5553.5\)- Denominator: \(140.6 + 165 = 305.6\)- Exponent: \(-\left[ \frac{5553.5}{305.6} \right] \approx -18.17\)- Finally: \(0.5 \times \exp(-18.17) \approx 6.39 \times 10^{-9}\).

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Key Concepts

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

Probability Approximation
Probability approximation is a useful tool when calculating probabilities directly is difficult. In cases where there's no formula for the standard normal cumulative distribution function (cdf), approximations can be relied upon. For standard normal distribution, one such useful approximation is detailed in the exercise we're examining. This specific approximation formula is given for when the variable is between 0 and 5.5:

\[ P(Z \geq z) \approx 0.5 \exp \left\{-\left[\frac{(83 z+351) z+562}{(703 / z)+165}\right]\right\}. \]

What does this mean? Well, the formula utilizes the exponential function to estimate the probability that a standard normal variable is greater than a particular value, denoted \(z\). The "exp" refers to the exponential function, which is pivotal in calculating probability distributions. Approximations like this simplify
  • calculation without a calculator or computer,
  • allow approximation within a slight margin of error.
For example, the formula demonstrates that for \(z=1\), the approximation is approximately 0.142, very close to what exact methods would indicate.
Normal Distribution Table
The Normal Distribution Table, or Z-table, is a vital resource in probability and statistics. It's used to find probabilities associated with the standard normal distribution, a continuous probability distribution that's symmetric and bell-shaped. In problems involving the standard normal distribution, we often need to look up probabilities or percentiles from this table.

The Z-table works because the standard normal distribution is standardized to have a mean of 0 and a standard deviation of 1. Each entry in the Z-table corresponds to a cumulative probability up to a given z-score. To find the probability that the variable falls within or beyond a certain z-score, you:
  • Identify the z-score of interest.
  • Navigate the table to locate the cumulative probability linked with that z-score.
  • Use this information to solve for the probability range you're interested in.
This educational tool transforms a potentially complicated computation into simply a lookup task. Comparing values found using an approximation to those in the table validates the accuracy of the former, as seen in our exercise.
Symmetry in Distributions
One of the key aspects of the standard normal distribution is its symmetry around 0. This means that the probability of a variable falling at a certain positive value from the mean is equal to the probability of it falling at the corresponding negative value. The area under the curve on either side of the mean is the same.

This symmetry is especially useful for simplifying calculations. For instance, when determining the probability of Z being less than -3, we can use symmetry to find it equals to 1 minus the probability of Z being greater than 3, since both probabilities are the same due to symmetry:
  • For example, \(P(Z < -3) = 1 - P(Z \geq 3)\).
  • Similarly, \(P(-a < Z < a) = 1 - [2 \times P(Z \geq a)]\).
Understanding symmetry helps in not just determining probabilities more efficiently but also in verifying your results, especially in symmetrical distribution scenarios.
Exponential Function in Probability
The exponential function is a crucial mathematical concept frequently used in probability, particularly in continuous distributions. It helps model various types of growth and decay processes, and its application is evident in probability approximation formulas too.

For standard normal distribution probabilities, the exponential function is part of the approximation. This function
  • Utilizes bases of mathematical constants,
  • Leverages natural occurrence in processes,
  • and simplifies into continuous data growth or decay rates.
In our exercise, the approximation formula employs an exponential decay model to give probability estimates, with the exponent calculated from the refined relationship:\[ -\left[\frac{(83 z+351) z+562}{(703 / z)+165}\right]. \]This makes exponential functions adaptable for more or less any smooth, continuous growth or decay scenario in probability calculations.

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Most popular questions from this chapter

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