95% of red lights last between 2.5 and 3.5 minutes.
<u>Step-by-step explanation:</u>
In this case,
- The mean M is 3 and
- The standard deviation SD is given as 0.25
Assume the bell shaped graph of normal distribution,
The center of the graph is mean which is 3 minutes.
We move one space to the right side of mean ⇒ M + SD
⇒ 3+0.25 = 3.25 minutes.
Again we move one more space to the right of mean ⇒ M + 2SD
⇒ 3 + (0.25×2) = 3.5 minutes.
Similarly,
Move one space to the left side of mean ⇒ M - SD
⇒ 3-0.25 = 2.75 minutes.
Again we move one more space to the left of mean ⇒ M - 2SD
⇒ 3 - (0.25×2) =2.5 minutes.
The questions asks to approximately what percent of red lights last between 2.5 and 3.5 minutes.
Notice 2.5 and 3.5 fall within 2 standard deviations, and that 95% of the data is within 2 standard deviations. (Refer to bell-shaped graph)
Therefore, the percent of red lights that last between 2.5 and 3.5 minutes is 95%
Answer: 0.0228
Step-by-step explanation:
Given : The mean and the standard deviation of finish times (in minutes) for this event are respectively as :-

If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.
Let X be the random variable that represents the finish times for this event.
z score : 

Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-

Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228
a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as
... c = k·s² + m·s + n
Filling in the given values gives three equations in k, m, and n.

Subtracting each equation from the one after gives

Subtracting the first of these equations from the second gives

Using the next previous equation, we can find m.

Then from the first equation
[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]
There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with
... c = 0.01s² - s + 37
b) At 150 kph, the cost is predicted to be
... c = 0.01·150² -150 +37 = 112 . . . cents/km
c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.
d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.
If their mean is 5, then the numbers add up to 25.
Here are a few sets that do the job:
1, 2, 3, 4, 15
10, 20, 30, 40, -75
+2, -2, +99, -99, 25
1, 5, 5, 5, 9