## order 3 years ago the lengths in cm of drinking straws produced in a factory have a normal distribution with mean mu and variance 0.64. it is given that 10% of the straws are shorter than 20cm (i) find the value of mu. It says you first find the s.d. which is sqrt:0.64 which is 0.8. then with the formula you place $20- mu \over 0.8$.Then it says <10% or >90%. The mark scheme uses z>90% so it ends up as (p) = -1.282 according to the normal distribution chart. So, you equate that formula to that, but why can't it be <10%?

1. kropot72

To use <10% you need a -z z table|dw:1337763325038:dw|

2. order

This doesn't help...

3. kropot72

z = -1.282$-1.282=\frac{20-\mu}{0.8}$ $-1.0256=20-\mu$ $\mu=21.0256cm$

4. order

$0.1<{20- mu \over 0.8}$ Why can't you use?

5. kropot72

$z \neq0.1$ The z-value is the standardised value of a random variable. If you look at the -z z table (which has negative values of z), you would find the z-value of -1.282 leads you to p = 0.1

6. order

Ok, It's true. Thanks :) I realized my mistake.

7. kropot72

You're welcome :)