Suppose "L" is a variable representing the volume of paint in a plastic tub.
"L" is normally distributed with mean "\\mu=10.25" and variance "\\sigma^2:L\\sim N(10.25, \\sigma^2)."
If "L\\sim(N, \\sigma^2)," then "Z={L-\\mu \\over \\sigma}\\sim N(0, 1)"
(a) Assume that "\\sigma^2=0.04"
Then
(b) Find the value of standard deviation so that 98% of tubs contain more than 10 liters of paint.
Given that
Then
"{L-\\mu \\over \\sigma}=Z^*""\\sigma={L-\\mu \\over Z^*}"
"\\sigma={10-10.25 \\over -2.053749}\\approx0.1217\\approx \\sqrt{0.0148}"
If the value of standard deviation is "0.1217," then 98% of tubs contain more than 10 liters of paint.
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