Determine the area of the normal distribution with mean of 10, standard deviation of 5 and scores between 5 to 12
We have a normal distribution, "\\mu=10, \\sigma=5."
Let's convert it to the standard normal distribution,
"z=\\cfrac{x-\\mu}{\\sigma}."
"z_1=\\cfrac{5-10}{5}=-1;\\\\\nz_2=\\cfrac{12-10}{5}=0.4;\\\\\nP(5<X<12)=P(-1<Z<0.4)=\\\\\n=P(Z<0.4)-P(Z<-1)=\\\\\n=0.6554-0.1587=0.4967 \\text{ (from z-table).}"
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