91影视

The probability of an event occurring is defined by probability. There are many scenarios in which we must forecast the outcome of an event in real life. The outcome of an event may be certain or uncertain. In such instances, we say that the event has a chance of happening or not happening.

Given

\(\begin{array}{l}p = 10\% = 0.1\\n = 1000\end{array}\)

(a) We are interested in the number of successes X (transmission errors) within the sample of 1000 bits, which then has a binomial distribution.

Requirements for a normal approximation of the binomial distribution:\(np \ge 5\;and\;nq \ge 5\).

\(\begin{array}{l}p = 1000(0.1) = 100 \ge 5\\nq = n(1 - p) = 1000(1 - 0.1) = 900 \ge 5\end{array}\)

Thus, the requirements are satisfied and we can then approximate the binomial distribution by the normal distribution.

The z-score is the value (using the continuity correction) decreased by the mean n p and divided by the standard deviation\(\sqrt {npq} = \sqrt {np(1 - p)} \).

\(z = \frac{{x - np}}{{\sqrt {np(1 - p)} }} = \frac{{125.5 - 1000(0.10)}}{{\sqrt {1000(0.10)(1 - 0.10)} }} \approx 2.69\)

Determine the corresponding normal probability using table.

\(\begin{array}{c}P(X \le 125) = P(X < 125.5)\\ = P(Z < 2.69)\\ = 0.9964\\ = 99.64\% \end{array}\)

(b) The distribution of the first transmission X and the distribution of the second transmission Y both have approximately normal distributions with mean

\(\mu = np = 1000(0.1) = 100\)

And standard deviation

\(\sigma = \sqrt {npq} = \sqrt {np(1 - p)} = \sqrt {1000(0.1)(1 - 0.1)} = \sqrt {90} = 3\sqrt {10} \approx 9.4868\).

Their difference X-Y then also has approximately a normal distribution with mean

\({\mu _{X - Y}} = {\mu _X} - {\mu _Y} = 100 - 100 = 0\)

and standard deviation:

\({\sigma _{X - Y}} = \sqrt {\sigma _X^2 + \sigma _Y^2} = \sqrt {{{(\sqrt {90} )}^2} + {{(\sqrt {90} )}^2}} = \sqrt {90 + 90} = \sqrt {180} = 6\sqrt 5 \approx 13.4164\)

The z-score is the value (using the continuity correction) decreased by the mean and divided by the standard deviation.

\(z = \frac{{x - {\mu _{X - Y}}}}{{{\sigma _{X - Y}}}} = \frac{{ \pm 50.5 - 0}}{{6\sqrt 5 }} \approx \pm 3.76\)

Determine the corresponding normal probability using the table.

\(\begin{array}{l}P(|X - Y| \le 50) = P(|X - Y| < 50.5) = P( - 50.5 < X - Y < 50.5) = P( - 3.76 < Z < 3.76)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = P(Z < 3.76) - P(Z < - 3.76) \approx 1 - 0 = 1\end{array}\)

Or using technology:

\(P(|X - Y| \le 50) = 0.999830 = 99.9830\% \)