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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$?$

(a) .6568 (b) 41.56 V (c) .3197

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

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were given a normal random variable with mean equals 40 and standard deviation equals 1.5. And for part a were asked for the probability of X being between 39 42. So if we standardize this question, it's the probability that 39 minus 40 over 1.5 is less than or equal to Zed, which is less than or equal to 40 to minus 40/1 400.5. And this probability comes out to 0.656 now for Part B were asked. For what? What is the value such that the probability of X exceeding it is only 15%. This is the 85th percentile. It is the value such that 85% of the distribution is less than it. So in terms of the standard normal distribution, the cumulative probability at this percent AL is 0.85 and that gives us a set value equal to 1.36 and then to convert back to ex, we take said, multiplied by the standard deviation and add the mean and this gives us 41.5. So 41.55 is the 85th percentile. And for part C, we're asked, if we take four independent draws from this distribution, this normal distribution, what is the probability that at least one of them exceeds 42? So let's first find the probability that an individual exceeds 42. So the probability of X exceeding 42. So that is the probability of one draw from the distribution exceeding 42. So let's call why the random variable that is three number that exceed 42 out of a sample of four. So why is a binomial distribution based on a sample size of four and a probability of success of zero point 0918? Now we're interested in the probability that why is at least one this is equal to one, minus the probability that why is zero now the probability of success is 0.918 Therefore, the probability of failure is 0.9082 So we need all four to be failures for white equals zero. So this probability is 0.90 a two to the exponents for, and this probability comes out 2.3197 So out of a sample of for the probability that at least one exceeds 42 is 420.3197

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