Answer:
The probability that a particular driver had exactly two speeding violations is 0.009.
Step-by-step explanation:
We are given that a sample of 2,000 licensed drivers revealed the following number of speeding violations;
<u>Number of Violations</u> <u>Number of Drivers</u>
0 1,910
1 46
2 18
3 12
4 9
5 or more <u> 5 </u>
<u>Total</u> <u> 2000 </u>
<u />
Now, the data means that 1,910 drivers had 0 speeding violations and so on.
Now, we have to find the probability that a particular driver had exactly two speeding violations, that means;
Number of drivers having exactly two speeding violations = 18
Total numbers of drivers = 2000
So, the required probability = 
= 
= <u>0.009</u>
The probability p of an orangecandy is 0.2. The sample size = 100.
The mean is given by:
The standard deviation is given by:
The answers are: Mean = 20. Standard deviation = 4.
It would have taken you 5:30 for each mile. 55/10 is 5.5 thats 5 and a half, half a minute is 30 seconds. so 5 minutes and 30 seconds
AB is divided into 8 equal parts and point C is 1 part FROM A TO B, so the ratio is 1:7, with C being 1/7 of the way. The ratio is k, found by writing the numerator of the ratio (1) over the sum of the numerator and denominator (1+7). So our k value is 1/8. Now we need to find the rise and the run (slope) of the points A and B.
. That gives us a rise of -4 and a run of 12. The coordinates of C are found in this formula:
![C(x,y)=[ x_{1} +k(run), y_{1} +k(rise)]](https://tex.z-dn.net/?f=C%28x%2Cy%29%3D%5B%20x_%7B1%7D%20%2Bk%28run%29%2C%20y_%7B1%7D%20%2Bk%28rise%29%5D)
. Filling in accordingly, we have
![C(x,y)=[-3+ \frac{1}{8}(12),9+ \frac{1}{8}(-4)]](https://tex.z-dn.net/?f=C%28x%2Cy%29%3D%5B-3%2B%20%5Cfrac%7B1%7D%7B8%7D%2812%29%2C9%2B%20%5Cfrac%7B1%7D%7B8%7D%28-4%29%5D%20%20)
which simplifies a bit to
. Finding common denominators and doing the math gives us that the coordinates of point C are
. There you go!
Answer:
The 95% of confidence intervals
(2.84 ,2.99)
Step-by-step explanation:
A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16
given small sample size n =20
sample mean x⁻ =2.92
sample standard deviation 'S' =0.16
level of significance ∝ = 0.95
The 95% of confidence intervals
the degrees of freedom γ=n-1 =20-1=19
t-table 2.093
(2.92-0.0748,2.92+0.0748)
(2.84 ,2.99)
Therefore the 95% of confidence intervals
(2.84 ,2.99)
