Stats Help!!?

Suppose we have a binomial distribution with n=40 trials and probability of success p=0.38. The random variable r is the number of successes in the 40 trials, and the random varialbe representing the proporion of successes is /\p (--that's that p hat looking thing where that "/\" symbol should be on top of the "p" pointing up) anyway, is /\p=r/n.





a). Compute P(0.30) %26lt; (or equal to) /\p %26lt;(or equal to) 0.40)





b. Compute P(/\p %26gt;(or equal to) 0.40).





c. If p= 0.12, can we approximate /\p by a normal distribtion? Explain? Note: approximate values except /\p-values to four decimal places.

Stats Help!!?
r ~ Bin(40, 0.38)





a) .3 %26lt;= r/n %26lt;= .4


.3n %26lt;= r %26lt;= .4n


12 %26lt;= r %26lt;= 16





answer is P[r = 12] + P[r = 13] + P[r=14] + P[r=15] + P[r=16]





where P[r=x] = n!/(x!(n-x)!) (p)^x (1-p)^(n-x) where n = 40, and p = 0.38





b) use the same sum but add all the values for r %26gt;=16 up to and including 40, or you can find it as 1 - P[ r %26lt; 16]





c) the rule of thumb for using the normal approximation to the binomial is if np %26gt; 10 and n(1-p) %26gt; 10. i.e. the number of expected success and the number of expected failures are both greater than ten.








Based on part c I think you can use the normal approximation to the binomial for parts a and b as well, but make sure that is what you are asked to do and that it is valid. Don't forget to use the continuity correction as well.
Reply:hey look it up at the binomial distribution table and try to calculate it

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