Answer:
a) 
b) 0.6957
Step-by-step explanation:
Let X represent the number of 4's when n = 5 independent spins
each has a probability of 0.2 (i.e p = 0.2)
This notation is represented as:
X
Binomial (n = 5, p = 0.2)
Probability of
number of 4's is:

here;
is the combinatorial expression
= 


So; let's first find:







b)
Given that:
The ratio of boys to girls at birth in Singapore is quite high at 1.09:1
What proportion of Singapore families with exactly 6 children will have at least 3 boys?
Probability of having a boy =
= 0.5215
Binomial Problem with n = 6
P(3<= x <=6) = 1 - P(0<= x <=2)
= 1 - binomial (6,0.5215,2)
= 0.6957
The average value of a continuous function f(x) over an interval [a, b] is
![\displaystyle f_{\mathrm{ave}[a,b]} = \frac1{b-a}\int_a^b f(x)\,dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Cmathrm%7Bave%7D%5Ba%2Cb%5D%7D%20%3D%20%5Cfrac1%7Bb-a%7D%5Cint_a%5Eb%20f%28x%29%5C%2Cdx)
We're given that
![\displaystyle f_{\rm ave[-1,2]} = \frac13 \int_{-1}^2 f(x) \, dx = -4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Crm%20ave%5B-1%2C2%5D%7D%20%3D%20%5Cfrac13%20%5Cint_%7B-1%7D%5E2%20f%28x%29%20%5C%2C%20dx%20%3D%20-4)
![\displaystyle f_{\rm ave[2,7]} = \frac15 \int_2^7 f(x) \, dx = 8](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Crm%20ave%5B2%2C7%5D%7D%20%3D%20%5Cfrac15%20%5Cint_2%5E7%20f%28x%29%20%5C%2C%20dx%20%3D%208)
and we want to determine
![\displaystyle f_{\rm ave[-1,7]} = \frac18 \int_{-1}^7 f(x) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Crm%20ave%5B-1%2C7%5D%7D%20%3D%20%5Cfrac18%20%5Cint_%7B-1%7D%5E7%20f%28x%29%20%5C%2C%20dx)
By the additive property of definite integration, we have

so it follows that
![\displaystyle f_{\rm ave[-1,7]} = \frac18 \left(\int_{-1}^2 f(x)\,dx + \int_2^7 f(x)\,dx\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Crm%20ave%5B-1%2C7%5D%7D%20%3D%20%5Cfrac18%20%5Cleft%28%5Cint_%7B-1%7D%5E2%20f%28x%29%5C%2Cdx%20%2B%20%5Cint_2%5E7%20f%28x%29%5C%2Cdx%5Cright%29)
![\displaystyle f_{\rm ave[-1,7]} = \frac18 \left(3\times(-4) + 5\times8\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Crm%20ave%5B-1%2C7%5D%7D%20%3D%20%5Cfrac18%20%5Cleft%283%5Ctimes%28-4%29%20%2B%205%5Ctimes8%5Cright%29)
![\displaystyle f_{\rm ave[-1,7]} = \boxed{\frac72}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Crm%20ave%5B-1%2C7%5D%7D%20%3D%20%5Cboxed%7B%5Cfrac72%7D)
Q1 is 2 2/3 or 2 2 out of 3 or 2.66666666667
Q5 is 3 1 out of 3 or 3 1/3 or 3.33333333
Q6 is 27/12 or 27 out of 12
Q10 is 109/10 or 109 out of 10
There are more compact cars (4*10 = 40) compared to trucks (2*10 = 20); however, the pictogram might make it appear that there are more trucks because the individual truck icon is larger compared to an individual compact car icon.
To anyone giving this image a quick glance, they may erroneously conclude that there are more trucks since their eye would notice the trucks first. Also, the person might think there are more trucks because bigger sizes tend to correspond to more proportion.
In real life, a truck is larger than a compact car, but the icons need to be the same size to have the figure not be misleading.
A very similar issue happens with the mid-size cars vs the compact cars as well. The three mid-size car icons span the same total width as the compact cars do, indicating that a reader might mistakenly conclude that there are the same number of mid-size cars compared to compact ones (when that's not true either).
Answer:
maybe you can attach the screenshot of the task?