Answer:
The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
Step-by-step explanation:
Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.
The random variable <em>X</em> is exponentially distributed with mean 7 minutes.
Then the parameter of the distribution is,
.
The probability density function of <em>X</em> is:
Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:
![=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7B%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7Be%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5B-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%5D%5E%7B9%7D_%7B6%7D%5C%5C%5C%5C%3De%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%206%7D-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%209%7D%5C%5C%5C%5C%3D0.424373-0.276453%5C%5C%5C%5C%3D0.14792%5C%5C%5C%5C%5Capprox%200.148)
Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
6y - 8 ≤ 10 Add 8 to both sides
6y ≤ 18 Divide both sides by 6
y ≤ 3
Y is less than or equal to 3.
Answer:
C. F(x) = (x+2)²
Step-by-step explanation:
A horizontal shift to the right by "h" units is achieved by replacing x with (x-h). The shift we want is to the left by 2 units, so h = -2, and we have ...
f(x) = g(x -(-2)) = g(x+2)
f(x) = (x+2)²
Answer:
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
d. z= 1.3322
Step-by-step explanation:
We formulate our hypothesis as
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
According to the given conditions
p`= 31/225= 0.1378
np`= 225 > 5
n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5
p = 0.4 x= 31 and n 225
c. Using the test statistic
z= p`- p / √pq/n
d. Putting the values
z= 0.1378- 0.11/ √0.11*0.89/225
z= 0.1378- 0.11/ √0.0979/225
z= 0.1378- 0.11/ 0.02085
z= 1.3322
at 5% significance level the z- value is ± 1.645 for one tailed test
The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept Ha : p >0.11 and conclude that the data indicates that the 11% of the world's population is left-handed.
The rejection region is attached.
The P- value is calculated by finding the corresponding value of the probability of z from the z - table and subtracting it from 1.
which appears to be 0.95 and subtracting from 1 gives 0.04998
It says that there was an error with my answer, but I don't know what, so I screenshotted what I typed and attached them as 4 images below:
