3. If a week is 7 days, what fraction of a week is 2 days?

In 2011, a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed sinc

coldgirl [10]

Answer:

Note: The full question is attached as picture below

a) Hо : p = 0.71

Ha : p ≠ 0.71

p = x / n

p = 91/110

p = 0.83.

1 - Pо = 1 - 0.71 = 0.29.

b) Test statistic = z

= p - Pо / [√Pо * (1 - Pо ) / n]

= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]

= 0.12 / 0.043265

= 2.77360453

Test statistic = 2.77

c) P-value

P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028

P-value = 0.0056

∝ = 0.01

P-value < ∝

Reject the null hypothesis. There is sufficient evidence to support the researchers claim at the 1% significance level.

6 0

3 years ago

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

If M is the midpoint of XY, find the coordinates of X if M(-3, -1) and Y(-8, 6) A. (2.5, -5.5) B. (-5.5, 2.5) C. (0, 0) D. (2, -

nadezda [96]

Answer:

(2, -8)

Step-by-step explanation:

(x, y)=(-3, -1)

(x1+x2)/2=-3,

(y2+y2)/2=-1,

-------------------

(x1+x2)/2=-3

(y1+y2)/2=-1

---------------------

x1-8=-6

x1=2

----------

y1+6=-2

y1=-8

(2, -8)

6 0

2 years ago

12 less the product of an unknown number and three equals -9.

chubhunter [2.5K]

In standard form its written
3x-12=-9
if you need help solving then here you go as well
3x-12=-9
add 12 to both sides
3x=3
divide both sides by 3
x=1

4 0

2 years ago

There are 20 cars in Thomas's toy car collection. Four-Fifths of the cars are blue. How many blue cars does he have?

Dmitry_Shevchenko [17]

Answer:

I think 15.

Step-by-step explanation:

20 ÷ 5=4 and 5 × 3=15.

8 0

2 years ago