Statistics questions part II?

Halp! hehe





1. Find an interval that covers the middle 95% of X-N(64,8).





2. Let X be a normal random variable with mean = 12 and SD = 2. Find the 10th percentile of this distribution.





3. The weight X of water melons is normally distributed with mean = 10 and SD = 2 pounds. Find c such that P(X%26gt;c) = .60





4. The annual rainfall X (in inches) at a certain region is normally distributed with mean = 40 inches and SD = 4. What is the probability that starting with this year, it will take more than 10 years before a year occurs having a rainfall of over 50 inches?





5. n=400; mean (u) = $8; SD = $20





a) The population from where the sample of 400 was selected does not follow the normal distribution. Why?





b) For what value of W can we say that P(u - W %26lt; Xbar %26lt; u + W) is equal to 80%?





c) Let T be the total overstatement for the 400 randomly selected items. Find the number b so that P(T %26gt; b) = .975

Statistics questions part II?
1. Find an interval that covers the middle 95% of X-N(64,8).





Middle interval = mean - 2sd -%26gt; mean +2sd


= 64 - 2sqr(8) to 64 +2sqr(8)





I'm assuming here that mean =64 and variance =8








2. Let X be a normal random variable with mean = 12 and SD = 2. Find the 10th percentile of this distribution.





Using inverse normal function,


P(x%26lt;c)=0.1 implies c= 9.44 to 2 dec pl








3. The weight X of water melons is normally distributed with mean = 10 and SD = 2 pounds. Find c such that P(X%26gt;c) = .60





P(X%26gt;c) = .60


implies P(X%26lt;c) = .40


implies c = 9.49 to 2 dec pl





4. The annual rainfall X (in inches) at a certain region is normally distributed with mean = 40 inches and SD = 4. What is the probability that starting with this year, it will take more than 10 years before a year occurs having a rainfall of over 50 inches?





Using normal cumulative probability,


p(X%26gt;50) = 0.0062


Let Y be the event, Number of years in which rainfall is greater than 50 in n years


Assuming this probability remains constant, you have a binomial distribution.


So you want the probability of getting zero occurences of event in 10 trials.


ie P(y=0) if y is binomial with n=10, p=0.0062


P(y=0) = 0.9397





ie there is a 94% probability that you won't have %26gt; 50 inches of rain in each of the next 1 years.





(I think thats how you would do the last one)