Answer:
The proportion of children that have an index of at least 110 is 0.0478.
Step-by-step explanation:
The given distribution has a mean of 90 and a standard deviation of 12.
Therefore mean,
= 90 and standard deviation,
= 12.
It is given to find the proportion of children having an index of at least 110.
We can take the variable to be analysed to be x = 110.
Therefore we have to find p(x < 110), which is left tailed.
Using the formula for z which is p( Z <
) we get p(Z <
= 1.67).
So we have to find p(Z ≥ 1.67) = 1 - p(Z < 1.67)
Using the Z - table we can calculate p(Z < 1.67) = 0.9522.
Therefore p(Z ≥ 1.67) = 1 - 0.9522 = 0.0478
Therefore the proportion of children that have an index of at least 110 is 0.0478
I know you have 1 and 3 right but I'm sorry about not knowing the rest.
Given:
8 2/3 hours per month
12 months in a year.
Number of hours she play soccer in a year.
8 2/3 must be converted to improper fraction.
((8*3)+2)/3 = 26/3
26/3 * 12 = (26 * 12)/3 = 312/3 = 104
Yolanda spends 104 hours in a year playing soccer.
Let the actual distance be x mile.
1/4 inch = 1/2 mile
9/2 inch = x mile
(9/2)/(1/4)=x/(1/2)
18=2x
x=9
the actual distance is 9 mile.
P(5,2) = n! /(n-r)!
n = 5, r = 2:
= (5 x 4 x 3 x 2 x 1 ) / 3 x 2 x 1
Cancel out common factors:
= 5 x 4 = 20
The answer is 20.