AP STATISTICS Random Variables

**asilvester635** · December 11th 2012, 08:19 PM

While he was a prisonerof war during World War II, John Kerrich tossed a coin10,000 times. He got 5067 heads. If the coin is perfectlybalanced, the probability of a head is 0.5.

a.) Is there reason to think that Kerrich's coin was notbalanced? To answer this question, use a Normaldistribution to estimate the probability that tossinga balanced coin 10,000 times would give a countof heads at least this far from 5000 (that is, at least5067 heads or no more than 4933 heads).

can anyone enlighten me and tell me what the question is asking???? would really appreciate it.. thank you

**abender** · December 14th 2012, 10:43 PM

$P\left(X\leq4933\right) + P\left(X \geq 5067\right)$

Note: By symmetry, $P\left(X\leq4933\right) = P\left(X \geq 5067\right)$ . However, I'll proceed with the individual calculations.

${\tiny P\left(X\leq 4933\right) = P\left(Z\leq\frac{4933-\mu}{\sigma}\right) = P\left(Z\leq\frac{4933-np}{\sqrt{npq}}\right) = P\left(Z\leq\frac{4933-10000\cdot0.5}{\sqrt{10000\cdot0.5^2}}\right) = P(Z\leq-1.34)=1-P(Z\leq1.34) \approx1-0.9099 = 0.0901}$

${\tiny P\left(X\geq 5067\right) = 1 - P\left(Z\leq\frac{5067-\mu}{\sigma}\right) = 1-P\left(Z\leq\frac{5067-np}{\sqrt{npq}}\right) = 1-P\left(Z\leq\frac{5067-10000\cdot0.5}{\sqrt{10000\cdot0.5^2}}\right) = 1-P(Z\leq1.34) \approx 1-0.9099 = 0.0901}$

Therefore, $P\left(X\leq4933\right) + P\left(X \geq 5067\right) \approx 2(0.0901)=0.1802$ .

Thread: AP STATISTICS Random Variables

LinkBack

Thread Tools

Display

AP STATISTICS Random Variables

Re: AP STATISTICS Random Variables

Similar Math Help Forum Discussions

Hello, I'm having problems with statistics problems with multiple random variables

What's the difference between discrete random variables and continous random variable

Order Statistics, N independent uniform random variables

Mathematical statistics: two-dimensional random variables

Statistics... Two Dimensional Random Variables

Search Tags