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Suppose that you are in a class of 31 students and it is assumed that approximately 13% of the population is left-handed. (Give...Asked by Nora 56
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
(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
Answers
Answered by
bobpursley
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.
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.
Answered by
Nora 56
I worked this out like you said above and got (a)9.936 (b)9.955 and both of them were wrong, any suggestions????
Answered by
jenna
where did you get .87 ?
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