Answer to Question #200873 in Statistics and Probability for Tashmi

Question

1.     Given the normally distributed variable X with mean 18 and variance 6.25. find

a)   P(X is less than15)

b)   THE VALUE OF K SUCH THAT P(X is less than k) = 0.2236

c)    P(17 less than X less than 21)

Answer 1

$The \space random\space variable \space X \space follows \space normal\ distribution \space\\ with \space a \space mean\space \mu \space =18 \space ,variance \space =6.25 \space and \space the \space standard\space deviation \space of \space \sigma=2.5\\ (a)\\ Lets \space find \space P(X<15)\\ First, \space we \space we\space have \space to \space find\space z\space value \space corresponding \space to \space x=15\\ z=\frac{x-\mu}{\sigma}=\frac{15-18}{2.5}=-1.2\\ hence\\ P(X<15)=P(Z<-1.2)=0.1151\\ put \space value \space from\space z \space table\\ (b)\\ Lets \space find \space the \space value\space of \space k\space such\space that \space P(X)=0.2236\\ P(z<\frac{k-\mu}{\sigma})=0.2236\\ P(z<-0.76)=0.2236\\ put \space value \space from\space z \space table\\ hence\\ \frac{k-\mu}{\sigma}=-0.76\\ k=-0.76\sigma +\mu\\ k=-0.76×2.5 +18\\ k=16.1\\ (c) \\ Lets\space find\space P(17<X<21)\\ first, \space we \space have \space to \space find\space z\space values\space corresponding \space to \space x_1=17 \space and \space x_2=21\\ z_1=\frac{x_1-\mu}{\sigma}=\frac{17-18}{2.5}=-0.4\\ z_2=\frac{x_2-\mu}{\sigma}=\frac{21-18}{2.5}=1.2\\ hence\\ P(17<X<21)=P(-0.4<Z<1.2)\\ =P(Z<1.2)-P(Z<-0.4) put \space value \space from\space z \space table\\ =0.8849-0.3446 =0.5403$