Normal Distribution problem.

Hello again. Here is the problem and my solution, if anyone can tell me if I am correct or cross-check our answers, I will be grateful!

Carrots grown by a particular farmer have weights in grammes approximately distributed as X, where X-Ν(75, (20)^2) (i.e X has a normal distribution with mean 75 grammes and standard deviation 20 grammes).

a) calculate the probability that a typical carrot weights between 65 and 80 grammes.

b) fin the probability that a bag of five carrots (whose weights can be assumed to be mutually independent) weight at least 350 grammes.

c) the farmer develops a new genetically modified carrot which has weight Y-N(110,σ^2) grammes. If the farmer wants the probability that the weight of a typical new carrot exceeds that of typical old carrot to be at least 0.9, what is the largest value that σ can be? (You may assume that the old and new carrot weights are independent).

This is my solution:

a) P(65<X<80)= P((65-75)/20<z<(80-75)/20)=P(-0.5<z<0.25)=P(z<0.25)-P(z<-0.5)=P(z<0.25)-1+P(z<0.5)=0.5987-1+0.6915=0.2902 (By statistical tables)

b) Let A=X1+X2+X3+X4+X5, where X1..X5 are the weights of the five carrots.

Then A-N(5*75,5*20^2) => A-N(375,2000)

Thus P(A>=350)=P(A>350)=P(z>(350-375)/sqrt(2000))=P(z>-0.56)

=1-P(z<=-0.56)=1-0.2877=0.7123 by statistical tables

c) Y-N(110,σ^2)

We want P(Y>75)>=0.9

=> 1-P(Y<=75)>=0.9

=> Φ((75-110)/σ))<=0.1

=> Φ(-35/σ)<= 0.1

=> -35/σ<= (-1.2)

=> σ<=29.16

The largest value of σ equals approximately to 29.16

If anyone can do the exercise and let me know or if I am correct or if I have been somewere wrong. Thanks in advance! Kind regards, apreciate your time!