How to Calculate the Probability of Extreme Weights in Production?

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The discussion focuses on calculating probabilities related to the weights of items produced by a production line, which are normally distributed with a mean of 12 ounces and a standard deviation of 2 ounces. The probability of a randomly selected item weighing between 8 and 16 ounces is approximately 0.9544, while the probability of an item weighing over 20 ounces is effectively 1. For quality control, the probability that 3 out of 7 randomly selected items meet the weight requirements is calculated using the binomial distribution, yielding a very low probability. The final challenge involves determining the probability that an item weighs either more than 14 ounces or less than 10 ounces, which can be solved by adding the probabilities of these mutually exclusive events. The discussion emphasizes the application of normal and binomial distribution formulas in solving these probability problems.
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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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