Answer:
a) 0.283 or 28.3%
b) 0.130 or 13%
c) 0.4 or 40%
d) 30.6 mm
Step-by-step explanation:
z-score of a single left atrial diameter value of healthy children can be calculated as:
z=
where
- X is the left atrial diameter value we are looking for its z-score
- M is the mean left atrial diameter of healthy children (26.7 mm)
- s is the standard deviation (4.7 mm)
Then
a) proportion of healthy children who have left atrial diameters less than 24 mm
=P(z<z*) where z* is the z-score of 24 mm
z*=
≈ −0.574
And P(z<−0.574)=0.283
b) proportion of healthy children who have left atrial diameters greater than 32 mm
= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm
z*=
≈ 1.128
1-P(z<1.128)=0.8703=0.130
c) proportion of healthy children have left atrial diameters between 25 and 30 mm
=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm
z(30)=
≈ 0.702
z(25)=
≈ −0.362
P(z<0.702)=0.7587
P(z<−0.362)=0.3587
Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4
d) to find the value for which only about 20% have a larger left atrial diameter, we assume
P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.
Then P(z<z*)=0.8 and z*=0.84. That is
0.84=
solving this equation for X we get X=30.648
We assume the lunch prices we observe are drawn from a normal distribution with true mean 
and standard deviation 0.68 in dollars.
We average 
samples to get 
.
The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write
Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains .
Our interval takes the form of 
as is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".
Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.
Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.
With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.
We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is
in other words a margin of error of
dollars
That's around plus or minus 17 cents.
Answer:
The answer is 3
Step-by-step explanation:
The answer is 3 because the number with x is always the slope.
Answer:
1. x > 4
2. m < 2
3. x > 2
4. x < -6
Step-by-step explanation:
These 4 inequalities will be solved the same way we solve equations. We take variables to one side and numbers to another and use algebra to solve. Each of them are solved shown below:
1.
so x is greater than 4, we can write (variable first):
x > 4
2.
so m is less than 2, we can write:
m < 2
3.
so x is greater than 2, we can write:
x > 2
4.
so x is less than -6, we can write:
x < -6
There are 3 times as many participants in the 40–59 age group than in the 0–19 age group is the one statement among the following choices given in the question that best compares the height of the bars of the histogram. The correct option would be " There are 3 times as many participants in the 40–59 age group than in the 0–19 age group."
