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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%?

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To use <10% you need a -z z table|dw:1337763325038:dw|
This doesn't help...
z = -1.282\[-1.282=\frac{20-\mu}{0.8}\] \[-1.0256=20-\mu\] \[\mu=21.0256cm\]

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\[0.1<{20- mu \over 0.8}\] Why can't you use?
\[z \neq0.1\] 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
Ok, It's true. Thanks :) I realized my mistake.
You're welcome :)

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