Statistics help!!!!!!?

Question

Lea has a pizza business. The times to prepare and deliver pizzas are aprox. normally disributed w/mean =20 mins & standard deviation of 10 mins.
a) Suppose that leo advertises "If it take more than 30 mins to get your pizza, you get it free." What frarction of this take-out-pizzas will he have to give away free?
b) how long a time should Leo allow if he wants to give away no more than 5% of the pizzas?

Answer 1

Z = (x - mean)/ sd = (30 - 20)/10 = 1

P(Z<1) = 0.8413 ===> P(Z>1) = 0.1587

a. 15.87 %

b. P(Z c = 1.655
x = 1.655*10 + 20 = 36.55 mins.

d:

Answer 2

Well, 30 mins =mean+1sd
a) 68% of the data falls within 1sd in a normal distribution, hence 32% doesn't. Half of this (16%) will be below the mean, so 84% of the pizzas should be delivered in under 30 minutes as 68+16=84.
Due to this, another 16% will be delivered after this time, and would be given away for free.
b) unless this is a very high level question, I wouldn't be surprised if it has been asked wrongly, as 5% is the amount that is outside 2sd's above AND below the mean. Assuming that this is a high level question, you have to work out how much below mean+2sd's this amount falls.
95% falls between 2sd's of the mean, but naturally, another 2.5% falls below the mean, outside of 2sd's.
To make this problem easier, we can pretend that the bottom half of the curve doesn't exist, so therefore 47.5% out of 50% is within 2sd's. Within 1sd, this would mean there was 34% (half of 68%). The difference between 47.5 and 34 is 13.5%. This means that as 2.5% of this is required it is over one fifth of the way between 47.5 and 34, in a reverse direction. Simply, this would make your answer about 38 minutes. However, as a normal distribution follows a bell shaped line, you can assume this figure as slightly higher, due to the constantly reducing rate of occurance of a delivery time. Put simply, this would mean your answer is around 39 minutes.
EDIT: Unsurprisingly, the answers you have got so far are all different, this is because it is almost impossible to accurately measure.
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Answer 3

a) the reduced variable to take 30 min is

z= t-mean/std

for 30 min z= 30-10/10 =1

you find in the table the probability 1-0.841= 0.159

b) in the table you find that to have less than 5% you must have z=1.6 so you must give a time 30+1.6*10 =46 minutes