While she was training for a marathon Lucy ran a total of 120 kilometers. How many meters did she run

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

3 years ago

You roll a number cube 20 times. The number 4 is ro

elena-14-01-66 [18.8K]

It would be 40 answer A

5 0

3 years ago

Felipe has 28 shirts to fold. Let N be the number of shirts he would have left to fold after folding F of them.

Debora [2.8K]

Ok let’s say he folds N=14 so F would be the other q4

8 0

3 years ago

Laila is 14 years older than tom. Laila is 5 years younger than Dan. Altogether they have a combined age of 51. How old are they

notka56 [123]

Laila=l
Tom=t
Dan=d

l=t+14
l=d-5
l+t+d=51

In terms of l:
d=l+5
t=l-14

So:
l+l+5+l-14=51
3l+5-14=51
3l-9=51
3l=60
l=20
d=25
t=6

Hope this helps :)

7 0

2 years ago

Find x, GH, and HI <br> .

ruslelena [56]

Given that GI = 53, and
• GH = 3x - 11
• HI = 2x + 4

We can establish the following equality statement to solve for x:

GH + HI = GI
3x - 11 + 2x + 4 = 53

Combine like terms:
5x - 7 = 53

Add 7 to both sides:
5x - 7 + 7 = 53 + 7
5x = 60

Divide both sides by 5 to solve for x:
5x/5 = 60/5

x = 12

Substitute the value of x into the equality statement to verify if it is the correct value for x:

GH + HI = GI
3x - 11 + 2x + 4 = 53

3(12) - 11 + 2(12) + 4 = 53
36 - 11 + 24 + 4 = 53
53 = 53 (True statement. Thus, x = 12 is the correct value).

Therefore:
GH = 3x - 11
GH = 3(12) - 11
GH = 25

HI = 2x + 4
HI = 2(12) + 4
HI = 28

Please mark my answers as the Brainliest, if you find this helpful. :)

6 0

2 years ago