Question

question 8
1 pts
the new york times reported that the mean time to download the homepage for the internal revenue service is 0.75 seconds. what is the probability that it takes you more than 1.2 seconds to download the page? (use an exponential distribution to determine this probability.)

use the following information to help you:
$e^{-0.1}=.9048$ $e^{-0.4}=.6703$ $e^{-0.8}=.4493$ $e^{-1.6}=.2019$ $e^{-3}=.0498$
$e^{-0.167}=.8465$ $e^{-0.417}=.6592$ $e^{-0.833}=.4346$ $e^{-1.667}=.1889$ $e^{-3.333}=.0357$
$e^{-0.2}=.8187$ $e^{-0.45}=.6376$ $e^{-0.9}=.4066$ $e^{-1.8}=.1653$ $e^{-3.75}=.0235$
$e^{-0.25}=.7788$ $e^{-0.5}=.6065$ $e^{-1}=.3679$ $e^{-2}=.1353$ $e^{-4}=.0183$
$e^{-0.267}=.7659$ $e^{-0.6}=.5488$ $e^{-1.2}=.3012$ $e^{-2.25}=.1054$ $e^{-5}=.0067$
$e^{-0.3}=.7408$ $e^{-0.625}=.5353$ $e^{-1.25}=.2865$ $e^{-2.4}=.0907$ $e^{-6}=.0024$
$e^{-0.333}=.7165$ $e^{-0.667}=.5134$ $e^{-1.333}=.2636$ $e^{-2.5}=.0821$ $e^{-8}=.0003$
$e^{-0.375}=.6873$ $e^{-0.75}=.4724$ $e^{-1.5}=.2231$ $e^{-2.667}=.0695$ $e^{-10}=.0000$

.3012
.2019
.6988
.5353
.7308
.7981

question 9
1 pts
on a december day, the probability of snow is.35. the probability of a frigid day is.60. the probability of snow and frigid weather is.20. snow and frigid weather are examples of:

dependent events

Explanation:

1/\mu$. So substituting $\lambda = 1/\mu$ into $P(X > x)$, we get $P(X > x) = e^{-x/\mu}$. Ah! There we go. So $P(X > x) = e^{-x/\mu}$. So $x = 1.2$, $\mu = 0.75$, so $x/\mu = 1.2 / 0.75 = 1.6$, so $P(X > 1.2) = e^{-1.6} = 0.2019$? But the table has $e^{-1.2}=.3012$. Wait, maybe the mean is 1.2? No, the problem says 0.75. Wait, maybe the question is using $\mu = 1.2$? No, the problem says 0.75. I'm confused. Wait, the options include.3012, which is $e^{-1.2}$. So maybe the mean is 1.2? But the problem says 0.75. Alternatively, maybe I made a mistake in the calculation of $x/\mu$. 1.2 divided by 0.75: 0.75 times 1.6 is 1.2, so 1.2 / 0.75 = 1.6. So $e^{-1.6}=0.2019$. So the answer should be.2019? But the option.3012 is $e^{-1.2}$. Wait, maybe the problem has a mistake, and the mean is 1.2. If the mean is 1.2, then $\mu = 1.2$, $\lambda = 1/1.2 = 5/6$, then $P(X > 1.2) = e^{-\lambda \times 1.2} = e^{-1} = 0.3679$, no. Wait, no. Alternatively, maybe the question is asking for $P(X < 1.2)$, which would be $1 - e^{-x/\mu} = 1 - e^{-1.6} = 1 - 0.2019 = 0.7981$, but that's an option. Wait, no, the question is "more than 1.2 seconds", so $P(X > 1.2)$. Wait, the options are.3012,.2019,.6988,.5353,.7308,.7981. Wait, 1 - 0.3012 = 0.6988. Oh! Wait, maybe I used the wrong formula. Maybe the cumulative distribution function is $F(x) = 1 - e^{-\lambda x}$, so $P(X > x) = e^{-\lambda x}$, but if $\lambda = 1/\mu$, then $P(X > x) = e^{-x/\mu}$. But maybe the question is using $\lambda = \mu$, which is incorrect, but let's check. If $\lambda = 0.75$, then $P(X > 1.2) = e^{-0.75 \times 1.2} = e

Answer:

1/\mu$. So substituting $\lambda = 1/\mu$ into $P(X > x)$, we get $P(X > x) = e^{-x/\mu}$. Ah! There we go. So $P(X > x) = e^{-x/\mu}$. So $x = 1.2$, $\mu = 0.75$, so $x/\mu = 1.2 / 0.75 = 1.6$, so $P(X > 1.2) = e^{-1.6} = 0.2019$? But the table has $e^{-1.2}=.3012$. Wait, maybe the mean is 1.2? No, the problem says 0.75. Wait, maybe the question is using $\mu = 1.2$? No, the problem says 0.75. I'm confused. Wait, the options include.3012, which is $e^{-1.2}$. So maybe the mean is 1.2? But the problem says 0.75. Alternatively, maybe I made a mistake in the calculation of $x/\mu$. 1.2 divided by 0.75: 0.75 times 1.6 is 1.2, so 1.2 / 0.75 = 1.6. So $e^{-1.6}=0.2019$. So the answer should be.2019? But the option.3012 is $e^{-1.2}$. Wait, maybe the problem has a mistake, and the mean is 1.2. If the mean is 1.2, then $\mu = 1.2$, $\lambda = 1/1.2 = 5/6$, then $P(X > 1.2) = e^{-\lambda \times 1.2} = e^{-1} = 0.3679$, no. Wait, no. Alternatively, maybe the question is asking for $P(X < 1.2)$, which would be $1 - e^{-x/\mu} = 1 - e^{-1.6} = 1 - 0.2019 = 0.7981$, but that's an option. Wait, no, the question is "more than 1.2 seconds", so $P(X > 1.2)$. Wait, the options are.3012,.2019,.6988,.5353,.7308,.7981. Wait, 1 - 0.3012 = 0.6988. Oh! Wait, maybe I used the wrong formula. Maybe the cumulative distribution function is $F(x) = 1 - e^{-\lambda x}$, so $P(X > x) = e^{-\lambda x}$, but if $\lambda = 1/\mu$, then $P(X > x) = e^{-x/\mu}$. But maybe the question is using $\lambda = \mu$, which is incorrect, but let's check. If $\lambda = 0.75$, then $P(X > 1.2) = e^{-0.75 \times 1.2} = e