Erica invests $10,000 at 5% interest compounded annually.

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

Please help me with number 8 please!!

forsale [732]

The answer to number eight is
That it would have to reflect in the middle and it has Proportional angles.

7 0

2 years ago

Height, in meters, is measured for each person in a sample. After the data are collected, all the height measurements

charle [14.2K]

Answer:

(E) The z-scores of the height measurements

Step-by-step explanation:

Mean is given by

$\bar {x}=\frac{1}{n}\left(\sum _{i=1}^{n}{x_{i}}\right)=\frac{x_{1}+x_{2}+\cdots +x_{n}}{n}$

Where,

n = Number of people

$x_{i}$ = The heights of people

When converting m to cm the mean would be

$\bar {x}=\frac{1}{n}\left(\sum _{i=1}^{n}{x_{i}}\right)\times 100$

So, the mean would change

The median gives us the middle data value when the data values are in ascending order.

So, the median would change

Standard deviation

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\bar{x})^2}$

When converting to centimeters

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}\left((x_i-\bar{x})\times 100\right)^2$

Hence, the standard deviation would change

When converting to centimeters the maximum height in meters would be the maximum height in centimeters also.

The z score is given by

$z=\frac{x-\mu}{\sigma}$

where,

x = Data point

$\mu$ = Mean

$\sigma$ = Standard deviation

When converting to centimeters

$z=\frac{(x-\mu)\times 100}{\sigma\times 100}\\\Rightarrow z=\frac{x-\mu}{\sigma}$

Hence, the z score would remain the same

5 0

3 years ago

A secret code consists of 3 digits, followed by 2 letters, followed by 1 of the following symbols: @

WINSTONCH [101]

The answer is B hope this helps

4 0

3 years ago

At the city museum, child admission is $5.20 and adult admission is $8.90 . on friday, 159 tickets were sold for a total sales o

Zinaida [17]

Total tickets sold = 159
Total sales = $1100.60
Child admission = $5.20
Adult admission = $8.90

Assumed all 159 tickets are Child admission tickets
Total sales = 159 x 5.2 = $826.80

Difference in amount = $1100.60 - $826.80 = $273.80
The difference must be contributed by the Adult Admission Tickets, which has a difference of $8.90 - $5.20 = $3.70

$273.80 ÷ $3.70 = 74 adult tickets

159 - 74 = 85 child tickets

5 0

3 years ago