Question

A.) for a binomial distribution with p= 0.65, find the smallest value of n so that the normal distribution is a reasonable approximation to the binomial distribution.

N=?

B.) given the Bernoulli experiment with n=60 and p= 0.6, use the normal distribution to estimate the following (round answers to four decimal places)

P(x=40)

P(x=28)

P(x=32)

C.) given the Bernoulli experiment with n= 12 and p= 0.5, use the normal distribution to estimate the following (round to four decimal places)

P(4 < x < 8)

P(4< or equal to X < or equal to 8)

P(7x< or equal to X < or equal to 8)

Answer 1

A) Given need to be satisfied so that the normal distribution is a reasonable approximation to the binomial distribution.

We are given p = 0.65

So,

$n(0.65)\geq 5$

$=>n\geq 5/0.65$

$=>n\geq 7.692\approx 8$

So, Smallest value of n so that the normal distribution is a reasonable approximation to the binomial distribution.

is n = 8

B)

$P\left (\frac{39.5-np}{\sqrt{np(1-p)}} <\frac{X-np}{\sqrt{np(1-p)}}<\frac{40.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left (\frac{39.5-60*0.6}{\sqrt{60*0.6(1-0.6)}} <\frac{X-60*0.6}{\sqrt{60*0.6(1-0.6)}}<\frac{40.5-60*0.6}{\sqrt{60*0.6(1-0.6)}} \right )$

$=P\left (0.922<Z<1.186 \right ) = P(Z<1.186)-P(Z<0.922)$

$P\left (\frac{27.5-np}{\sqrt{np(1-p)}} <\frac{X-np}{\sqrt{np(1-p)}}<\frac{28.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left (\frac{27.5-60*0.6}{\sqrt{60*0.6(1-0.6)}} <\frac{X-60*0.6}{\sqrt{60*0.6(1-0.6)}}<\frac{28.5-60*0.6}{\sqrt{60*0.6(1-0.6)}} \right )$

$=P\left (-2.24<Z<-1.9764 \right ) = P(Z<-1.9764)-P(Z<-2.24)$

$P\left (\frac{31.5-np}{\sqrt{np(1-p)}} <\frac{X-np}{\sqrt{np(1-p)}}<\frac{32.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left (\frac{31.5-60*0.6}{\sqrt{60*0.6(1-0.6)}} <\frac{X-60*0.6}{\sqrt{60*0.6(1-0.6)}}<\frac{32.5-60*0.6}{\sqrt{60*0.6(1-0.6)}} \right )$

$=P\left (-1.186<Z<-0.922 \right ) = P(Z<-0.922)-P(Z<-1.186)$

C)

$=P\left (\frac{4.5-np}{\sqrt{np(1-p)}} <\frac{X-np}{\sqrt{np(1-p)}}<\frac{7.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left (\frac{4.5-12*0.5}{\sqrt{12*0.5(1-0.5)}} <\frac{X-12*0.5}{\sqrt{12*0.5(1-0.5)}}<\frac{7.5-12*0.5}{\sqrt{12*0.5(1-0.5)}} \right )$

$=P\left (-0.866<Z<0.866 \right ) = P(Z<0.866)-P(Z<-0.866)$

$P(4\leq X\leq 8)= P(3.5<X<8.5)$

$=P\left (\frac{3.5-np}{\sqrt{np(1-p)}} <\frac{X-np}{\sqrt{np(1-p)}}<\frac{8.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left (\frac{3.5-12*0.5}{\sqrt{12*0.5(1-0.5)}} <\frac{X-12*0.5}{\sqrt{12*0.5(1-0.5)}}<\frac{8.5-12*0.5}{\sqrt{12*0.5(1-0.5)}} \right )$

$=P\left (-1.443<Z<1.443 \right ) = P(Z<1.443)-P(Z<-1.443)$

$P(7\leq X\leq 8)= P(6.5<X<8.5)$

$=P\left (\frac{6.5-np}{\sqrt{np(1-p)}} <\frac{X-np}{\sqrt{np(1-p)}}<\frac{8.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left (\frac{6.5-12*0.5}{\sqrt{12*0.5(1-0.5)}} <\frac{X-12*0.5}{\sqrt{12*0.5(1-0.5)}}<\frac{8.5-12*0.5}{\sqrt{12*0.5(1-0.5)}} \right )$

$=P\left (0.2886<Z<1.443 \right ) = P(Z<1.443)-P(Z<0.2886)$