Question

(a) During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. what is the probability that 6 particles enter the counter in a given milliseconds
(b) The height of adult women in the United States is normally distributed with mean 64.5 inches and S.D 2.4 inches. Find the probability that a randomly chosen women is a) less than 63 inches tall b) less than 70 inches tall c) between 63 and 70 inches tall. d) Alice is 72 inches tall. What percentage of women is shorter than Alice

Answer 1

Answer on Question #45892 – Math – Statistics and Probability

Question.

(a) During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. What is a probability that 6 particles enter the counter in a given millisecond?

(b) The height of adult women in the United States is normally distributed with mean 64.5 inches and S.D 2.4 inches. Find the probability that a randomly chosen women is a) less than 63 inches tall b) less than 70 inches tall c) between 63 and 70 inches tall. d) Alice is 72 inches tall. What percentage of women is shorter than Alice?

Solution.

(a) Assume that the particle stream is the easiest. Then the number $\xi$ of radioactive particles passing through a counter in 1 millisecond has the Poisson's distribution, i.e.

$$P(\xi = k) = \frac{\lambda^k}{k!} e^{-\lambda} \text{ where } k = 0, 12, \dots, \text{ and by the hypothesis } \lambda = 4. \text{ So},$$ $$P(\xi = 6) = \frac{4^6}{6!} e^{-4} \approx 0.104.$$

(b) Let $\xi$ be the height of adult women in the United States. Then $\eta = \frac{\xi - 64.5}{2.4}$ has a standard normal distribution, i.e. $P(\eta < x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-\frac{u^2}{2}} du$. The tabulated function of Laplace has the form $\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{u^2}{2}} du$.

a) $P(\xi < 63) = P\left(\frac{\xi - 64.5}{2.4} < \frac{63 - 64.5}{2.4}\right) = P\left(\frac{\xi - 64.5}{2.4} < -0.625\right) = 0.5 - \Phi(0.625) = 0.5 - 0.23565 = 0.26435$.

b) $P(\xi < 70) = P\left(\frac{\xi - 64.5}{2.4} < \frac{70 - 64.5}{2.4}\right) = P\left(\frac{\xi - 64.5}{2.4} < 2.29\right) = 0.5 + \Phi(2.29) = 0.5 + 0.48928 = 0.98928$.

c) $P(63 < \xi < 70) = P\left(\frac{63 - 64.5}{2.4} < \frac{\xi - 64.5}{2.4} < \frac{70 - 64.5}{2.4}\right) = P\left(-0.625 < \frac{\xi - 64.5}{2.4} < 2.29\right) = \Phi(0.625) + \Phi(2.29) = 0.23565 + 0.48928 = 0.72493$.

d) $P(\xi < 72) = P\left(\frac{\xi - 64.5}{2.4} < \frac{72 - 64.5}{2.4}\right) = P\left(\frac{\xi - 64.5}{2.4} < 3.125\right) = 0.5 + \Phi(3.125) = 0.5 + 0.49903 = 0.99903 = 99.903\%$.

Answer. (a) 0.104

(b):

a) 0.26435

b) 0.98928

c) 0.72493

d) 99.903%

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