(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

the weight of items produced by a production line isnormally 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!)) * [tex]\pi[/tex]^k * (1-[tex]\pi[/tex])^(n-k)

mean = n[tex]\pi[/tex]

variance = n[tex]\pi[/tex](1-[tex]\pi[/tex])

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

3. The attempt at a solution

a. x~n (µ=12, [tex]\sigma[/tex]=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, [tex]\sigma[/tex]=.95)

p(x=3)

p(x=K)=[tex](3!)/(3!)(7-3)![/tex](.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, [tex]\sigma[/tex]=2)

P(x<10) or P(x>14)

Hope you people can help me

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# Need help with statistics

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