Stats q pls help

Question

Weights of erasers produced by a certain factory are known to follow the uniform distribution between 31.5 g and 32.3 g. (a) (10 points) Erasers produced by this factory are sold in packs of 45. A retailer randomly bought 200 packs. Find the probability that, for at least 15 packs, the average weight of the erasers in the pack is at least 31.95 g

Answer 1

We are told the distribution of weights of individual erasers: Uniform from 31.5 to 32.3 g. We'll call that random variable U, and its mean is the average (31.5+32.3) / 2 = 31.9g, and its variance is

|32.3-31.5| / 12g = 0.8/12 = 1 / 15 g^2 √which is 0.0667g^2. So U is Uniform(m=31.9g, sd=√0.0667g)

The distribution of weights of 45-packs (random variable V) is the distribution of the sum of 45 uniform's from 31.5 to 32.3. We will assume that the weights of those 45 erasers are independent. This sum of 45 independent variables is approximately normally distributed with mean 45 x 31.9 = 1,436g = 1.436kg, and its variance is 45 x (1 / 15) g^2 , so the standard deviation of V is √3 g or 1.732g .

We are interested in the event of a pack having average eraser weight 31.95 grams or more, which is the same event as the sum of the weights in the pack being 45 x 31.95g = 1,438g. or more. Now 1,438 is 2 grams above average for a pack, which is 2 / 1.732 = 1.155 standard deviations (V's sd's). According to Excel's standard normal cumulative probability function is 1 - NORM.S.DIST(1.155,TRUE) = 1-0.876 = 0.124 .

So the probability that a pack of erasers having an average weight of 31.95 grams or more is 0.124 .

The probability of that happening 15 or more times out of 200 draws is the probability of getting 15 or more successes in 200 Bernoulli trials with probability of success in each trial equal to p = 0.124. B(200,0.124) is approximately normally distributed with mean 200 x .124 and variance 200 x 0.124 x (1-0.124) = 21.7, so with standard deviation √21.7 = 4.66 trials.15 successes is 15 / 4.66 = 3.22 standard deviations.

The probability that approximately normal variable (number of Packs having weight 31.95 or more among 200 randomly chosen packs) being 3.22 or more standard deviations above average is about

1-NORM.S.DIST(3.22,TRUE) = 1 - = 0.000641 .