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Chapter 4: Q52E (page 169)

Let X denote the number of flaws along a \({\bf{100}}\)-m reel of magnetic tape (an integer-valued variable). Suppose X has approximately a normal distribution with m \(\mu = 25\) and s \(\sigma = 5\). Use the continuity correction to calculate the probability that the number of flaws is

a. Between \({\bf{20}}\) and \({\bf{30}}\), inclusive.

b. At most \({\bf{30}}\). Less than \({\bf{30}}\).

Short Answer

Expert verified

(a) The probability is \(P(a < Z < b) = \phi (b) - \phi (a)\).

(b)The probability is \(0.8643\).

Step by step solution

01

Introduction

When a discrete distribution is approximated by a continuous distribution, a continuity correction is applied.

02

Given Information

The number of faults along a\(100\)-meter reel is denoted by the symbol\(X\). It is assumed that\(X\)is normally distributed, with the following mean and standard deviation:

\(\mu = 25\)

\(\sigma = 5\)

03

Finding Probability

(a)

Due to the correction of continuity:

\(P(20 \le X \le 30) = P(20 - 0.5 \le X \le 30 + 0.5)\)

\( = P(19.5 \le X \le 30.5)\)

Standardizing gives:\(19.5 \le X \le 30.5\)if and only if

\(\frac{{19.5 - 25}}{5}{\rm{ }} \le \frac{{X - 25}}{5} \le \frac{{30.5 - 25}}{5}\)

\(\frac{{ - 5.5}}{5}{\rm{ }} \le \frac{{X - 25}}{5} \le \frac{{5.5}}{5}\)

\( - 1.1 \le Z \le 1.1\)

Thus

\(P(19.5 \le X \le 30.5){\rm{ }} = P( - 1.1 \le Z \le 1.1)\)

\( = \phi (1.1) - \phi ( - 1.1)\)

\( = 0.8643 - 0.1357\)

\(P(19.5 \le X \le 30.5){\rm{ }} = 0.7286\)

Let Z be a continuous rv with cdf as a proposition\(\phi (z)\). Then for any\(a\)and\(b\)with\(a < b\),

\(P(a < Z < b) = \phi (b) - \phi (a)\)

Therefore, the probability is \(P(a < Z < b) = \phi (b) - \phi (a)\).

04

Finding Probability

(b)

Due to the correction of continuity\(P(X \le 30)\)now becomes:

\(P(X \le 30) = P(X \le 30 + 0.5)\)

\( = P(X \le 30.5)\)

Standardizing gives:\(X \le 30.5\)if and only if

\(\frac{{X - 25}}{5} \le \frac{{30.5 - 25}}{5}\)

\(\frac{{X - 25}}{5} \le \frac{{5.5}}{5}\)

\( - Z \le 1.1\)

Thus

\(P(X \le 30.5) = P(Z \le 1.1)\)\( = \phi (1.1)\;\;\;{\rm{ use appendix A - 3 here }})\)

\( = 0.8643\)

Therefore, the probability is \(0.8643\).

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