Answer:
Step-by-step explanation:
The question is incomplete because the data is missing, i.e. the probability that you will score 5, 4, 3, 2, 1.
But it is resolved as follows:
[tex]P(x\geq 3) = P(\frac{x - m}{\frac{sd}{\sqrt{n} } } \geq \frac{3 - m}{\frac{sd}{\sqrt{n} }})\\\\[/tex]
where m is the mean and sd is the standard deviation.
the m is calculated by the sum of the multiplication of the score by the probability of this
that is to say,
score probability
5 0.2
4 0.3
3 0.1
2 0.3
1 0.1
m = 5*0.2 + 4*0.3 + 3*0.1 + 2*0.3 + 1*0.1
m = 3.2
However, the standard deviation will be calculated by
sd = [tex]\sqrt{\\}[/tex]∑[tex](x - m)^{2}*p[/tex]
that is, knowing the mean already, we can calculate the standard deviation, following the example:
sd =[tex]\sqrt{[(5-3.2)^2] *0.2 + [(4-3.2)^2] *0.3 + [(3-3.2)^2] *0.1 + [(2-3.2)^2] *0.3 + [(1-3.2)^2] *0.1 }[/tex]
sd = [tex]\sqrt{1.76}[/tex]
sd = 1.327
And also n = 5, because it's 5 scores. We replace in the initial equation:
[tex]P(x\geq 3) = P(Z \geq \frac{3 - 3.2}{\frac{1.327}{\sqrt{5} }})\\\\[/tex]
[tex]P(x\geq 3) = P(Z \geq -0.337)\\\\\\[/tex]
Therefore for the example the number z is -0.337, which if in the normal distribution table corresponds to 0.3520, that is the probability that the average is at least 3, for the example is 35.20 %.
