In a normal distribution, 69% of the distribution is less than 28 and 90% is less than 35. Find the mean
and standard deviation of the distribution
We are given that,
"p(Z\\lt {28-\\mu\\over \\sigma})=0.69" and "p(Z\\lt {35-\\mu\\over \\sigma})=0.90"
So,
"p(Z\\lt {28-\\mu\\over \\sigma})=0.69\\implies \\phi(0.69)={28-\\mu\\over \\sigma}". From the standard normal tables, we have that,
"\\phi(0.69)=0.4958503\\implies {28-\\mu\\over \\sigma}=0.4958503....(1)"
and
"p(Z\\lt {35-\\mu\\over \\sigma})=0.90\\implies \\phi(0.90)={35-\\mu\\over \\sigma}". From the standard normal tables, we have that "\\phi(0.90)=1.281552" . This implies that,
"{35-\\mu\\over \\sigma}=1.281552.......(2)"
We shall find the values of "\\mu" and "\\sigma" by solving equations 1 and 2.
From equation 1,
"{28-\\mu\\over \\sigma}=0.4958503\\implies 28-\\mu=0.4958503\\sigma\\implies \\sigma={28-\\mu\\over 0.4958503}.....(3)"
Substituting for the value of "\\sigma" from equation 3 in equation 2 we get,
"0.4958503(35-\\mu)=1.281552(28-\\mu)"
Simplifying this we have,
"17.3547605-0.4958503\\mu=35.883456-1.281552\\mu\\implies0.7857017\\mu=18.5286955\\implies \\mu=23.5823538" From equation 3, we can find the value of "\\sigma" as,
"\\sigma={28-23.5823538\\over 0.4958503}=8.90923369"
Therefore, the values for "\\mu" and "\\sigma" are 23.58 and 8.91 respectively. Both rounded off to 2 decimal places.
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