1 ) Billy walks 15 miles every day to soccer practice and he had 5 practices last week.
We know that, 1 mile = 1760 yards
So, 15 miles = (15× 1760) yards = 26400 yards
Thus the total distance walked in 5 practices will be:
2) Mark's current age is years
So, his age was in 10 years ago
Given that, his current age is 3 times his age 10 years ago.
So, the equation to represent this situation will be :
Answer:
a) Figure attached
b)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 95%
c)
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%
d)
And using the z score we got:
And that correspond to approximately 0.15%
Step-by-step explanation:
Part a
For this case we can see the figure attached.
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part b
Let X the random variable that represent the length of human pregnancy of a population, and for this case we know that:
Where and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 95%
Part c
And we can find this probability with this difference:
An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%
Part d
We want this probability:
And using the z score we got:
And that correspond to approximately 0.15%
