the value of k such that P(X<k) = 0.2578
Full problem:
Given the normally distributed variable X with mean 18 and standard deviation 2.5, find
(a) P(X<15):
(h) P(17< X<21);
(c) the value of k such that P(X<k) = 0.2578;
(d) the value of k such that P(X> A) = 0.1539.
"\\mu=18 \\\\\n\n\\sigma=2.5"
(a)
"P(X<15) = P(Z< \\frac{15-18}{2.5}) \\\\\n\n=P(Z< -1.2) \\\\\n\n= 0.1150"
(b)
"P(17<X<21) = P(X<21) -P(X<17) \\\\\n\n=P(Z< \\frac{21-18}{2.5}) -P(Z< \\frac{17-18}{2.5}) \\\\\n\n=P(Z< 0.8571) -P(Z< -0.4) \\\\\n\n= 0.8042 -0.3445 \\\\\n\n= 0.4597"
(c)
"P(X<k) = 0.2578 \\\\\n\nP(Z< \\frac{k-18}{2.5}) =0.2578 \\\\\n\n\\frac{k-18}{2.5} = -0.65 \\\\\n\nk-18 = 2.5 \\times (-0.65) \\\\\n\nk=18 -1.625 \\\\\n\nk=16.375"
(d)
"P(X>k) = 0.1539 \\\\\n\nP(X<k) = 1 -0.1539 = 0.8461 \\\\\n\nP(Z< \\frac{k-18}{2.5}) =0.8461 \\\\\n\n\\frac{k-18}{2.5} = 1.02 \\\\\n\nk-18=2.5 \\times 1.02 \\\\\n\nk = 18 +2.55 \\\\\n\nk=20.55"
Comments
Leave a comment