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State the Central Limit Theorem

The time spent by a factory worker A packing a box is a random variable with mean 3.5 minutes and standard deviation 1 minute. The times spent packing any two boxes are independent.

(i) What is the approximate probability that he packs 100 boxes in less than 6 hours?

(ii) What is the approximate probability that in 5.5 hours he will pack at least 90 and less than 100 boxes?

(iii) How long will he need to work for there to be 95% probability of packing 100 boxes?

(iv) A second worker B, does a similar job and the time it takes him to pack a box is a random variable with mean 3 minutes and standard deviation 1.2 minutes, independent of the time taken by A. The times spent by B packing any two boxes are independent. The two workers are assigned the job of packing 130 boxes each. Find the approximate probability that worker B will finish at least 50 minutes before worker A.

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Solution provides detailed explanation, along with equations, on how to calculate central limit theorem.

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i) Central limit theorem

Let X_1,X_2,...,X_N be a set of N independent random variates and each X_i have an arbitrary probability distribution P(x_1,...,x_N) with mean mu_i and a finite variance sigma_i^2. Then the normal form variate
X_(norm)=(sum_(i==1)^(N)x_i-sum_(i==1)^(N)mu_i)/(sqrt(sum_(i==1)^(N)sigma_i^2)) (1)

has a limiting cumulative distribution function which approaches a normal distribution.

ii) Denote by X1,X2,...,X100 the life of the corresponding bulb, respectively. Then we know that
X1,X2,...,X100 are independent and follow the same distribution with mean mu=2000 hrs and standard deviation sigma=250hrs. Then the average life of the selected bulbs is
...

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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