Normal distribution (mean, standard deviation), using a calculator to solve (2 of 2)

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At a soft drinks factory, a bottle filling machine fills an average of 10000 bottles a day with a standard deviation of 1000. Assuming that the production is normally distributed and that there are 260 working days in a year, calculate the approximate working days per year on which:

a) the machine fills at most 9000 bottles.

b) the machine fills at least 8000 bottles.

c) the machine fills between 9000 and 12000 bottles.

Answer:

Answer:

a) 9000 bottles are exactly within 1 standard deviation from the mean. We know that, in a normal distribution, between $latex \mu – \sigma$ and $latex \mu$ lies 34.13% of the data so the probability that the machine fills at most 9000 bottles is 100% – 50% – 34.13% = 15.87%.

This means that out of 260 working days in a year, the machine will fill at most 9000 bottles on approximately 41 days.

b) 8000 bottles are exactly within 2 standard deviations from the mean. We know that between $latex \mu – 2\sigma$ and $latex \mu$ lies 47.72% of the data so the probability that the machine fills at least 8000 bottles is 50% (the half on the other side of the mean) + 47.72% = 97.72%.

This means that out of 260 working days in a year, the machine will fill at least 8000 bottles on approximately 254 days.

c) Following the same reasoning, there is a 68.26% chance that the machine fills between 9000 and 11000 bottles per working day (because the average is 10000 and both 9000 and 11000 are within 1 standard deviation from it) and a 13.59% chance that the machine fills between 11001 and 12000 bottles per working day.

Therefore, the probability we’re looking for is 81.85%, which means the machine will fill between 9000 and 12000 bottles on approximately 213 working days in a year.

Next up:
Inverse normal