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  • Suppose that 90 % of people are right -handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Q. 4  Suppose that 90^{o}/_{o} of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

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best_answer

90^{o}/_{o} of people are right-handed.

P(right-handed)=\frac{9}{10}

P(left-handed)=q=1-\frac{9}{10}=\frac{1}{10}

at most 6 of a random sample of 10 people are right-handed.

the probability that more than  6 of a random sample of 10 people are right-handed is given by,

                                                        \sum_{T}^{10} ^{10}C_r P^{r} q^{10-r}

                                                       =\sum_{T}^{10} ^{10}C_r\frac{9}{10}^r .(\frac{1}{10})^{10-r}

the probability that at most 6 of a random sample of 10 people are right-handed is given by

                                                      =1-\sum_{T}^{10} ^{10}C_r .\frac{9}{10}^r .(\frac{1}{10})^{10-r}

Posted by

seema garhwal

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