How to Calculate the Probability of Extreme Weights in Production?

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SUMMARY

The discussion focuses on calculating probabilities related to the weights of items produced by a production line, which follows a normal distribution with a mean (µ) of 12 ounces and a standard deviation (σ) of 2 ounces. Key calculations include determining the probability of an item weighing between 8 and 16 ounces, which is approximately 95.44%, and the probability of an item weighing over 20 ounces, which is effectively 1. The discussion also addresses a binomial probability scenario involving quality control for 7 items, where the probability of exactly 3 items meeting the weight requirement is calculated. Lastly, participants explore how to find the probability of an item weighing greater than 14 ounces or less than 10 ounces, emphasizing the additive nature of these mutually exclusive events.

PREREQUISITES
  • Understanding of normal distribution (N(µ, σ))
  • Familiarity with binomial distribution and related formulas
  • Knowledge of z-scores and their application in probability calculations
  • Basic statistical concepts such as mean and standard deviation
NEXT STEPS
  • Learn how to calculate z-scores for normal distribution problems
  • Study binomial probability calculations in-depth
  • Explore the Central Limit Theorem and its implications for sample distributions
  • Investigate statistical software tools like R or Python for probability simulations
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Statisticians, data analysts, quality control engineers, and students studying probability and statistics will benefit from this discussion.

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Homework Statement


the weight of items produced by a production line is normally distributedwith a mean of 12 ounces and a standard deviation of 2 ounces.
a. what is the probability that a randomly selected item will weight between 8 and 16 ounces? (DONE)
b. what is the probability that a randomly selected item will weight over 20 ounces? (DONE)
c. Suppose that quality control requires the weight of items to be within 8 and 16 ounces. You select 7 items at random (each item is independent). What is the probability that 3 of the items will fulfill quality control requirements. (DONE BUT HAVE DOUBTS)
d. find the probability that a randomly selected item has a weight that is greater than 14 or smaller than 10. (STUCK IN THIS ONE)

2. Related formulas
if x is BIN
p(x=k) = (n!)/((n-k)!(k!)) * \pi^k * (1-\pi)^(n-k)
mean = n\pi
variance = n\pi(1-\pi)

if x is N(µ,\sigma), then z=(x-µ)/(\sigma) is N(0,1)

The Attempt at a Solution


a. x~n (µ=12, \sigma=2)
p(8<x<16)
p(x<16) - p(x<8)
z=(16-12)/2 z=(8-12)/2
z=2 z= -2
p(z<2)-p(z<-2)
=.9772-.0228
=.9544

b.P(x>20)
z=(20-12)/2
z=4
p(z>4)= 1

c.x~binomial (µ=7, \sigma=.95)
p(x=3)
p(x=K)=(3!)/(3!)(7-3)!(.95)^3 (1-.95)^4
.
.
.
x=0.000187551

d. I have no idea how to deal with this one
I think I have to use the mean and standard deviation of the problem
(µ=12, \sigma=2)
P(x<10) or P(x>14)
Hope you people can help me
 
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619snake said:

Homework Statement


the weight of items produced by a production line is normally distributedwith a mean of 12 ounces and a standard deviation of 2 ounces.
a. what is the probability that a randomly selected item will weight between 8 and 16 ounces? (DONE)
b. what is the probability that a randomly selected item will weight over 20 ounces? (DONE)
c. Suppose that quality control requires the weight of items to be within 8 and 16 ounces. You select 7 items at random (each item is independent). What is the probability that 3 of the items will fulfill quality control requirements. (DONE BUT HAVE DOUBTS)
d. find the probability that a randomly selected item has a weight that is greater than 14 or smaller than 10. (STUCK IN THIS ONE)

2. Related formulas
if x is BIN
p(x=k) = (n!)/((n-k)!(k!)) * \pi^k * (1-\pi)^(n-k)
mean = n\pi
variance = n\pi(1-\pi)

if x is N(µ,\sigma), then z=(x-µ)/(\sigma) is N(0,1)

The Attempt at a Solution


a. x~n (µ=12, \sigma=2)
p(8<x<16)
p(x<16) - p(x<8)
z=(16-12)/2 z=(8-12)/2
z=2 z= -2
p(z<2)-p(z<-2)
=.9772-.0228
=.9544

b.P(x>20)
z=(20-12)/2
z=4
p(z>4)= 1

c.x~binomial (µ=7, \sigma=.95)
p(x=3)
p(x=K)=(3!)/(3!)(7-3)!(.95)^3 (1-.95)^4
.
.
.
x=0.000187551

d. I have no idea how to deal with this one
I think I have to use the mean and standard deviation of the problem
(µ=12, \sigma=2)
P(x<10) or P(x>14)
Hope you people can help me

The events X>14 and X<10 are mutually exclusive (cannot occur at the same time) so that their probabilities are additive:

P(X>14 or X<10) = P(X>14) + P(X<10)
 

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