Distributions help

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.

Reply 1

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

Reply 2

DazzaWazza

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!

Reply 3

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.