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The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value \(40 \mathrm{~V}\) and standard deviation \(1.5 \mathrm{~V}\). a. What is the probability that the voltage of a single diode is between 39 and 42 ? b. What value is such that only \(15 \%\) of all diodes have voltages exceeding that value? c. If four diodes are independently selected, what is the probability that at least one has a voltage exceeding 42 ?

Short Answer

Expert verified
a. 0.6568. b. 41.554 V. c. 0.3169.

Step by step solution

01

Identify the distribution parameters

First, identify the parameters of the normal distribution for the diodes. The mean voltage \( \mu \) is 40 V, and the standard deviation \( \sigma \) is 1.5 V.
02

Calculate probability for a single diode (Question a)

We need to find the probability that the voltage \( Z \) for a single diode is between 39 V and 42 V. We achieve this by calculating the z-scores. The z-score is calculated with the formula: \( z = \frac{X - \mu}{\sigma} \). For 39 volts: \[ z_1 = \frac{39 - 40}{1.5} = -\frac{1}{1.5} = -0.67 \]For 42 volts: \[ z_2 = \frac{42 - 40}{1.5} = \frac{2}{1.5} = 1.33 \]Using a standard normal distribution table or calculator, find the probability: \[ P(39 < Z < 42) = P(Z < 1.33) - P(Z < -0.67) \]\( P(Z < 1.33) \approx 0.9082 \), \( P(Z < -0.67) \approx 0.2514 \) Therefore, \[ P(39 < Z < 42) = 0.9082 - 0.2514 = 0.6568 \]
03

Find the cutoff voltage for exceeding (Question b)

We want to find the voltage value \( X \) such that the probability that a diode has a voltage exceeding \( X \) is 0.15, i.e., \( P(X > V) = 0.15 \). First, find the z-score where the cumulative probability is 0.85 (since \(1 - 0.15 = 0.85\)). This z-score corresponds to the 85th percentile:Using a z-table or calculator, \( z \approx 1.036 \) for which \( P(Z < z) = 0.85 \).Convert the z-score back to the voltage value: \[ V = \mu + z \times \sigma = 40 + 1.036 \times 1.5 = 40 + 1.554 = 41.554 \text{ volts} \]
04

Calculate probability for four diodes (Question c)

For four independently selected diodes, we need to find the probability that at least one diode exceeds 42 V. Let \( p \) be the probability that a single diode exceeds 42 V: Find \( p \): \[ P(Z > 1.33) = 1 - P(Z < 1.33) = 1 - 0.9082 = 0.0918 \]The probability that none of the four diodes exceeds 42 V is \( (1 - 0.0918)^4 \).Therefore, \[ P(\text{at least one diode exceeds 42 V}) = 1 - (0.9082)^4 = 1 - 0.6831 = 0.3169 \]

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Key Concepts

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

Probability
Probability is the measure of how likely an event is to happen. In the context of the normal distribution, it helps us determine the likelihood of a random variable falling within a specific range. When dealing with diodes and their breakdown voltage, we often calculate probabilities related to these voltage values.

To determine probabilities, we utilize the properties of the normal distribution. For instance, to find the probability of a diode's voltage being between 39 V and 42 V, we calculate the area under the normal distribution curve between these values. This involves determining z-scores for these voltages, as these scores help standardize the values and make them comparable against a standard normal distribution table.
  • Probability of voltage between 39 V and 42 V is calculated as the difference between the cumulative probabilities at these points.
  • For example, if the cumulative probability of 39 V is 0.2514 and for 42 V is 0.9082, the probability is the difference: 0.9082 - 0.2514 = 0.6568.
Z-Score
The z-score is a statistical measure that describes a value's relation to the mean of a group of values. It's expressed in terms of standard deviations. For any normal distribution, the z-score tells us where our value lies in comparison to the mean.

To calculate a z-score, use the formula: \[z = \frac{X - \mu}{\sigma}\]where \(X\) is the observed value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation of the distribution. In our diode example, if the mean voltage is 40 V and standard deviation is 1.5 V, a voltage of 39 V translates to a z-score of -0.67, indicating it is 0.67 standard deviations below the mean.
  • Z-scores allow us to use standard normal distribution tables to find probabilities for specific value ranges.
  • A positive z-score indicates the value is above the mean, while a negative indicates it's below.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In essence, it indicates how much the values in a data set deviate from the mean. A smaller standard deviation means the values are closer to the mean, while a larger one means they are more spread out.

In the case of diode voltages, the standard deviation of 1.5 V provides an idea of the range we might expect for the voltages around the mean of 40 V. The standard deviation is crucial for calculating the z-score, which subsequently is used to determine probabilities and understand the distribution of diode voltages.
  • It is important in risk assessment; smaller standard deviation implies less risk in voltage performance.
  • Helps in understanding the normal distribution curve's width. A larger standard deviation results in a flatter curve.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a particular value. In other words, it is the probability that a value will fall within a certain range in a normal distribution. This concept is crucial when working with normally distributed variables because it allows us to determine the likelihood of observing a value below a certain threshold.

Using cumulative probability tables, or a calculator, allows us to determine these thresholds for different z-scores. For instance, when finding a value such that only 15% of all diodes exceed it, we look for a cumulative probability of 0.85. This critical insight employs the *inverse* of the probability of exceeding a value (1 - cumulative probability).
  • Cumulative probabilities give insight into percentiles, like determining the 85th percentile.
  • They are pivotal in fields like quality control and risk management.

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