Let [latex]X=[/latex] the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or [latex]\frac{1}{4}[/latex] hour.)

[latex]x=0,1,2,3,...[/latex]

If Leah receives, on the average, six telephone calls in two hours, and there are eight 15-minute intervals in two hours, then Leah receives [latex]\left(\frac{1}{8}\right)\left(6\right)=0.75[/latex] calls in 15 minutes, on average. So, [latex]\mu=0.75[/latex] for this problem.

[latex]X{\sim}P(0.75)[/latex]

Find [latex]P(x>1)[/latex]. [latex]P(x>1)=0.1734[/latex] (calculator or computer)

• Press 1 – and then press 2^nd DISTR.
• Arrow down to poissoncdf. Press ENTER.
• Enter (.75,1).
• The result is [latex]P(x>1)=0.1734[/latex].

Note: The TI calculators use [latex]\lambda[/latex] (lambda) for the mean.

The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:

[latex]P(x>1)=1−\text{poissoncdf}(0.75,1)[/latex]

The graph of [latex]X{\sim}P(0.75)[/latex] is:

The [latex]y=[/latex]-axis contains the probability of [latex]x=[/latex] where [latex]X=[/latex] the number of calls in 15 minutes.

1. [latex]P(x=160)=\text{poissonpdf}(147, 160){\approx}0.0180[/latex]
2. [latex]P(x\leq160)=\text{poissoncdf}(147, 160){\approx}0.8666[/latex]
3. Standard Deviation[latex]=\sigma=\sqrt{\mu}=\sqrt{144}\approx12.1244[/latex]

Let [latex]X[/latex] = the number of days with low seismic activity.

Using the binomial distribution:

• [latex]P(x=10)[/latex] = binompdf(200, .0102, 10) ≈ 0.000039

Using the Poisson distribution:

• Calculate μ = np = 200(0.0102) ≈ 2.04
• [latex]P(x=10)[/latex] = poissonpdf(2.04, 10) ≈ 0.000045

We expect the approximation to be good because [latex]n[/latex] is large (greater than 20) and [latex]p[/latex] is small (less than 0.05). The results are close—both probabilities reported are almost 0.