Question

use the fact that the variance of a poisson distribution is $\sigma^2 = \mu$
the mean number of bankruptcies filed per hour by businesses in a country was about seven. (a) find the variance and the standard deviation. interpret the results. (b) find the probability that at most five businesses will file bankruptcy in any given hour.

(a) the variance is 7.000
(round to three decimal places as needed.)
the standard deviation is 2.6
(round to one decimal place as needed.)
choose the correct answer below.
a. the number of bankruptcies differs from the standard deviation by no more than the mean number of bankruptcies.
b. the number of bankruptcies always differs from the mean by no more than the standard deviation number of bankruptcies.
c. the number of bankruptcies is always the standard deviation number of bankruptcies.
d. the number of bankruptcies usually differs from the mean by no more than the standard deviation number of bankruptcies.

(b) $p$(at most five) = $\square$
(round to three decimal place as needed.)

Explanation:

Step1: Recall Poisson Probability Formula

The probability mass function for a Poisson distribution is \( P(X = k)=\frac{\mu^{k}e^{-\mu}}{k!} \), where \( \mu = 7 \) (mean number of bankruptcies per hour), \( k \) is the number of events, and \( e\approx2.71828 \). We need to find \( P(X\leq5)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5) \).

Step2: Calculate Each Term

• For \( k = 0 \): \( P(X = 0)=\frac{7^{0}e^{-7}}{0!}=e^{-7}\approx\frac{1}{1096.633}\approx0.000912 \)
• For \( k = 1 \): \( P(X = 1)=\frac{7^{1}e^{-7}}{1!}=7e^{-7}\approx7\times0.000912 = 0.006384 \)
• For \( k = 2 \): \( P(X = 2)=\frac{7^{2}e^{-7}}{2!}=\frac{49e^{-7}}{2}\approx\frac{49\times0.000912}{2}\approx0.022344 \)
• For \( k = 3 \): \( P(X = 3)=\frac{7^{3}e^{-7}}{3!}=\frac{343e^{-7}}{6}\approx\frac{343\times0.000912}{6}\approx0.052136 \)
• For \( k = 4 \): \( P(X = 4)=\frac{7^{4}e^{-7}}{4!}=\frac{2401e^{-7}}{24}\approx\frac{2401\times0.000912}{24}\approx0.091238 \)
• For \( k = 5 \): \( P(X = 5)=\frac{7^{5}e^{-7}}{5!}=\frac{16807e^{-7}}{120}\approx\frac{16807\times0.000912}{120}\approx0.127733 \)

Step3: Sum the Terms

\( P(X\leq5)=0.000912 + 0.006384+0.022344 + 0.052136+0.091238+0.127733 \)
\( = 0.000912+0.006384 = 0.007296 \)
\( 0.007296+0.022344 = 0.02964 \)
\( 0.02964+0.052136 = 0.081776 \)
\( 0.081776+0.091238 = 0.173014 \)
\( 0.173014+0.127733 = 0.300747\approx0.301 \) (rounded to three decimal places)

Answer:

\( 0.301 \)