Posted by
**Nora 56** on
.

Posted by Nora 56 on Sunday, June 2, 2013 at 7:32pm.

Sorry to repost but no one answered after I posted on the 2nd,please help,

Suppose that you are in a class of 31 students and it is assumed that approximately 13% of the population is left-handed. (Give your answers correct to three decimal places.)

(a) Compute the probability that exactly five students are left-handed.

Answer .161

(b) Compute the probability that at most four students are left-handed.

Answer .129

(c) Compute the probability that at least six students are left-handed.

Answer .193

•Math check answer - bobpursley, Sunday, June 2, 2013 at 8:19pm

a. what is .13^5 * .87^(31-5) >

Put .13^5 * .87^(31-5)= in your google search window.

b. at most four students...

add the probabliliy of one, two, three, four, and none are left handed.

Pr=.13^0*.87^31+.13^1*.87^30 + ...

c. at least six?

that is the same as 1- probability5orless

= 1- Pr(a)-Pr (b) where pr(a), pr(b) is in part a, and b.

•Math check answer - Nora 56, Thursday, June 6, 2013 at 10:01am

I worked this out like you said above and got (a)9.936 (b)9.955 and both of them were wrong, any suggestions????