/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Let \(X\) denote the number of f... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) denote the number of flaws along a \(100-\mathrm{m}\) reel of magnetic tape (an integer-valued variable). Suppose \(X\) has approximately a normal distribution with \(\mu=25\) and \(\sigma=5\). Use the continuity correction to calculate the probability that the number of flaws is a. Between 20 and 30 , inclusive. b. At most 30 . Less than 30 .

Short Answer

Expert verified
a) 0.7286 b) At most 30: 0.8643, Less than 30: 0.8159

Step by step solution

01

Understanding the Problem

We are given a normal distribution with mean \( \mu = 25 \) and standard deviation \( \sigma = 5 \). The task is to find probabilities for certain intervals of the variable \( X \), which represents the number of flaws in a magnetic tape reel. Because \( X \) is discrete, we will use the continuity correction for accurate probability calculations.
02

Identify the Correct Interval for Part a

The problem asks for the probability that the number of flaws is between 20 and 30, inclusive. To apply the continuity correction, we consider the interval 19.5 to 30.5. This adjustment accounts for the discrete nature of \( X \).
03

Convert the Adjusted Interval to Z-scores for Part a

We calculate the Z-scores for the continuity-corrected interval. The Z-score formula is given by \( Z = \frac{X - \mu}{\sigma} \).For 19.5: \( Z = \frac{19.5 - 25}{5} = -1.1 \)For 30.5: \( Z = \frac{30.5 - 25}{5} = 1.1 \)
04

Find Probability Using the Standard Normal Distribution for Part a

Using standard normal distribution tables or a calculator, find the probabilities:- \( P(Z < 1.1) \approx 0.8643 \)- \( P(Z < -1.1) \approx 0.1357 \)Therefore, the probability that \( X \) is between 19.5 and 30.5 is:\( P(-1.1 < Z < 1.1) = 0.8643 - 0.1357 = 0.7286 \)
05

Determine the Interval for Part b (At most 30)

For the probability that the number of flaws is at most 30, we use the interval ending at 30.5 for the continuity correction.
06

Convert to Z-score for Part b (At most 30)

Calculate the Z-score for 30.5:\( Z = \frac{30.5 - 25}{5} = 1.1 \)
07

Find Probability for Part b (At most 30)

Using the standard normal distribution table or a calculator:- \( P(Z < 1.1) \approx 0.8643 \)Thus, the probability that \( X \) is at most 30 is approximately 0.8643.
08

Determine the Interval for Part b (Less than 30)

For the probability that the number of flaws is less than 30, we use the interval up to 29.5 due to continuity correction.
09

Convert to Z-score for Part b (Less than 30)

Calculate the Z-score for 29.5:\( Z = \frac{29.5 - 25}{5} = 0.9 \)
10

Find Probability for Part b (Less than 30)

Using the standard normal distribution table or a calculator:- \( P(Z < 0.9) \approx 0.8159 \)Thus, the probability that \( X \) is less than 30 is approximately 0.8159.

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Key Concepts

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

Continuity Correction
Continuity correction is an essential concept when using the normal distribution to approximate probabilities for discrete variables. When a variable, like the number of flaws on a magnetic tape, can only take integer values, direct application of the normal distribution might lead to inaccuracies. This is because the normal distribution is continuous by nature. Therefore, to get precise probabilities, we make a slight adjustment.
For instance, instead of simply considering the interval from 20 to 30, we adjust it to 19.5 to 30.5. This accounts for the way discrete data make the jump from one value to another without passing through continuum values. By applying this correction, we better simulate the discrete nature of the data when using continuous models. - Helps bridge discrete and continuous. - Adjusts the calculation range subtly but significantly. - Ensures more accurate probability calculations.
Z-score Calculation
Z-score calculation is a crucial step in using the standard normal distribution to find probabilities. The Z-score represents how many standard deviations a particular value is from the mean of the distribution.
The formula used is:\[ Z = \frac{X - \mu}{\sigma} \]With \( \mu \) being the mean and \( \sigma \) the standard deviation. In our case, for an interval like 19.5 to 30.5 (from continuity correction), Z-scores help translate these bounds into a standard format. For instance, 19.5 becomes:\( Z = \frac{19.5 - 25}{5} = -1.1 \)And 30.5 turns into:\( Z = \frac{30.5 - 25}{5} = 1.1 \)- Standardizes scores regardless of distribution units.- Facilitates use of standard normal tables or calculators.- Helps directly compare different ranges.
Probability Calculation
Probability calculation forms the final piece of the puzzle after making continuity corrections and calculating Z-scores. Once we have Z-scores, we turn to the standard normal distribution to find the probabilities of interest.
The standard normal distribution, centered at 0 with a standard deviation of 1, allows us to assess the likelihood of a Z-score being within a particular range. Using Z-scores, we can determine probabilities such as:For Z-scores of -1.1 to 1.1:- \( P(-1.1 < Z < 1.1) = P(Z < 1.1) - P(Z < -1.1) \approx 0.8643 - 0.1357 = 0.7286 \)This indicates the probability that the number of flaws is between 20 and 30, inclusive.For a Z-score of up to 1.1 (interpreting "at most 30"):- \( P(Z < 1.1) \approx 0.8643 \)By following these steps, we derive probabilities that align closely with the discrete nature of underlying data, thanks to the continuity correction. This approach is pivotal in making educated decisions based on statistical analyses.

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