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# Distributions help watch

1. Could anyone help me with this;

Strips of metal are cut to length L cm where L~N(µ,0.5²).

a.) Given that 2.5% of the cut lengths exceed 50.98cm , show that µ=50.

P(L>50.98)=0.025
P(Z>(50.98-µ)/0.5)=0.025

What do i do from here...

b.) Find P(49.25<L<50.75). -Done-

c.) Those strips with length either less than 49.25cm or greater than 50.75cm cannot be used.

Two strips of metal are selected at random.

Find the probability that both strips cannot be used.
2. a) Find the value of Z in the table that is closest to 0.025, then let this equal (50.98-µ)/0.5

c) Probability one can't be used is 1 - P(49.25<L<50.75), then just square this value
3. (Original post by Bezza)
a) Find the value of Z in the table that is closest to 0.025, then let this equal (50.98-µ)/0.5

c) Probability one can't be used is 1 - P(49.25<L<50.75), then just square this value
Oh,

Thanks for the help!
4. For part a)

P(L > 50.98) = 0.025
1 - P(L < 50.98) = 0.025
P(L < 50.98) = 0.975
P(Z < [ 50.98 - µ ]/0.5 ) = 0.975

Now, from the normal distribution tables phi(1.96) = P(Z < 1.96) = 0.975. Therefore, comparing with our above equation we have

[ 50.98 - µ ]/0.5 = 1.96

Rearranging gives µ = 50.
5. (Original post by mikesgt2)
For part a)

P(L > 50.98) = 0.025
1 - P(L < 50.98) = 0.025
P(L < 50.98) = 0.975
P(Z < [ 50.98 - µ ]/0.5 ) = 0.975

Now, from the normal distribution tables phi(1.96) = P(Z < 1.96) = 0.975. Therefore, comparing with our above equation we have

[ 50.98 - µ ]/0.5 = 1.96

Rearranging gives µ = 50.
I feel so stupid.

Thanks for the help!

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