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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%?
 one year ago
 one year 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%?
 one year ago
 one year ago

This Question is Closed

kropot72Best ResponseYou've already chosen the best response.1
To use <10% you need a z z tabledw:1337763325038:dw
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
z = 1.282\[1.282=\frac{20\mu}{0.8}\] \[1.0256=20\mu\] \[\mu=21.0256cm\]
 one year ago

orderBest ResponseYou've already chosen the best response.0
\[0.1<{20 mu \over 0.8}\] Why can't you use?
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
\[z \neq0.1\] The zvalue 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 zvalue of 1.282 leads you to p = 0.1
 one year ago

orderBest ResponseYou've already chosen the best response.0
Ok, It's true. Thanks :) I realized my mistake.
 one year ago
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