Suppose the life span of an electrical component is exponentially distributed wi

Question

Suppose the life span of an electrical component is exponentially distributed with parameter lambda = 0.014. Define X_1,X_2,..., X_100 to be a simple random sample of 100 components" life spans. (a) What is the expected value of the average life span of the sample. E(X)'? (b) What is the minimum probability that the average life span of the sample will be within 15.2 units of the population mean? (c) What minimum sample size would be required in order that the minimum probability that the average life span of the sample will be within 15.2 units of the population mean is at least 0.98? (d) Suppose you obtained many samples of the minimum size you found in part (c). Describe how you expect the averages of these samples to be distributed, including the type of distribution, mean, and variance.

Explanation / Answer

(a) As we take sample random samples of 100 compnents life span. we can approximate that the mean value of these random sample shall follow normal distribution according to Central Limit thereom which has

Mean = Expected value of mean of the given distribution = Population mean = 1/ = 1/0.014 =  71.43 units

Standard deviation = Standard deviation of distribution/ sqrt (n) = (1/) / sqrt (100) = 71.43/ 10 = 7.143 units

So we have to find the minimum probbability that the sample mean will be under 15.2 units of population mean

Z = 15.2/ 7.143 = 2.128

so P - value = 1 - 2* Pr ( Z > 2.128), we multplied it will 2 to get results from either side distance of 15.2 units

P - value = 1- 2 * 0.01667 = 0.967 so there are 96.7 % probability there that average life span is within 15.2 units.

(c) Here Pr vaalue = 0.98

so respective Z - value from Z - table => Z = 2.326

so Z = 15.2 / ((1/) / sqrt (n) = 2.326

71.43/ sqrt(n) = 15.2/ 2.326 = 6.535

sqrt (n) = 10.93

n = 120

(d) Yes, THe CLT will be applied here and we expect the type of distribution shall be a normal one, Its mean will bethe mean of exponential disribution equals to 71.43 units.

Variance = (1/2 ) /n = (71.43)2 / 120 = 42.52 units

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