Thread: Normal Distribution Problem!-S-1!!

Hi... umm guys m stuck on this question... can sumbdy plz guide me!!

The time taken by a garage to replace worn-out brake pads follows a normal distribution with mean 90 minutes and standard deviation 5.8 minutes. The garage claims to complete the replacements in 'a to b minutes'. If this claim is to be correct for 90% of the repairs, find a and b correct to 2 s.f., bassed on a symmetrical interval centred on the mean. (ans: a=80 and b=100)

Can sumbdy plz help asap!!

It is known that if Z has a normal distribution with mean μ and standard deviation σ, then (Z - μ) / σ has the standard normal distribution, i.e., the one with mean 0 and standard deviation 1. We need to find Δ such that P(μ - Δ ≤ Z ≤ μ + Δ) = .9. Verify that μ - Δ ≤ Z ≤ μ + Δ is equivalent to -Δ / σ ≤ (Z - μ) / σ ≤ Δ / σ.

Since the standard normal distribution is symmetric, P(-Δ / σ ≤ (Z - μ) / σ ≤ Δ / σ) = 2P(0 ≤ (Z - μ) / σ ≤ Δ / σ). Now use the table such as this one to find Δ / σ such that 2P(0 ≤ (Z - μ) / σ ≤ Δ / σ) = .9. Since σ is given, you can find Δ.

Ohk. Thanks a lot. I was not trying to put the whole equation in terms of delta. Because of that i ended up in a single equation connecting a and b. Thanks a bunch.

I m stuck at another one :/

A market sells potatoes whose weights are normally distributed with mean 65 grams and standard deviation 15 grams.
(i) Find the probability that a randomly chosen potato weighs between 40 grams and 80 grams.

I found the answer to this part using the probability distribution table and all and the answer was 0.7935 which is correct, however I can't figure out the second part which says :

The market sells potatoes weighing more than 80 grams separately packaged. Potatoes weighing between 80 grams and L grams are labelled "large" and potatoes weighing over L grams are labelled as "extra large".
(ii) Given that a randomly chosen potato is twice as likely to be large as extra large ,calculate the value of L

Can sumbuddy plz guide me!

Originally Posted by IceDancer91

The market sells potatoes weighing more than 80 grams separately packaged. Potatoes weighing between 80 grams and L grams are labelled "large" and potatoes weighing over L grams are labelled as "extra large".
(ii) Given that a randomly chosen potato is twice as likely to be large as extra large ,calculate the value of L

Let X be the weight of a random potato. Then P(80 ≤ X ≤ L) = 2P(L ≤ X). Rewrite this equation so that it uses only P(X ≤ c) for various c. Then rewrite it in terms of P((X - 65) / 15 ≤ ...) and use the fact that (X - 65) / 15 has the standard normal distribution.

Ohh ohk. Thanks.