Hi

i need help with the attached question

I have attempted the question but i am stuck on parts d) to h)

My answers so far are

a) i substituted in $\displaystyle z=3$ hence my answer was 0.112

b)$\displaystyle P(X>2\mid X<3)=\frac{0.184}{0.888}$=0.207

c)$\displaystyle X\sim Bin(6,\:0.112) P(X=2)=\dbinom{6}{2}(0.112)^{2}(1-0.112)^{4}$=0.117

my guess to part d was

$\displaystyle \dbinom{6}{0}(0.112)^{0}(0.888)^{6}$=0.4903

Probability of first three not failing is $\displaystyle (0.888)^{3}$=0.7 hence answer is $\displaystyle \frac{0.4903}{0.7}$=0.7

any help is appreichated

thanks