1. The problem statement, all variables and given/known data The time Rafa spends on his homework each day is normally distributed with mean 1.9hrs and standard deviation σ. On 80% of these days he spends more than 1.35 hours on his homework. i. find the value of σ ii. find the probability that Rafa spends less than 2 hours on his homework iii. A random sample of 200 days is taken. Use an approximation to find the probability that the number of days Rafa spends more than 1.35hrs on homework is between 163days and 173days inclusive 2. Relevant equations 3. The attempt at a solution i. i have no problem here 80per cent=0.842 z=-0.842 -0.842=(1.35-1.9)/σ σ=0.653 ii. here i have a problem i agree with book that ##Pz=(2-1.9)/0.653## = P(z is less than 0.1531) now book says that this is equivalent to 0.561 (on one tailed z table) yet i am getting 0.5+0.1531= 0.6531 which is equivalent to 0.394 (on 1 tailed z distribution table). iii. text says x→normal distribution with (mean=160,standard deviation=32), my problem is how did they get σ=32?, i am assuming for 160, they did 80%times 200 the rest of the working that results to p=(0.9915-0.6707)=0.321 is ok with me.