ANSWER:
a. 9.9, 1.8
b. 9.9, 0.2882
c. 0.5239
d. 0.6368
e. No
f.
Q1 = 9.7069
Q3 = 10.0931
IQR = 0.3862
STEP-BY-STEP EXPLANATION:
a.
X ~ N (9.9, 1.8)
b.
x ~ N (9.9, 1.8/√39)
x ~ N (9.9, 0.2882)
c.
P(X > 9.8)
We calculate the probability as follows:
[tex]\begin{gathered} P\left(X>9.8\right)=1-p\left(\frac{X-9.9}{1.8}<\frac{9.8-9.9}{1.8}\right) \\ \\ P\left(X>9.8\right)=1-p(z<-0.06) \\ \\ P\left(X>9.8\right)=1-0.4761 \\ \\ P\left(X>9.8\right)=0.5239 \end{gathered}[/tex]
d.
p (x > 9.8)
We calculate the probability as follows:
[tex]\begin{gathered} P\left(x>9.8\right)=1-p\left(\frac{X-9.9}{\frac{1.8}{\sqrt{39}}}<\frac{9.8-9.9}{\frac{1.8}{\sqrt{39}}}\right) \\ \\ P\left(x>9.8\right)=1-p(z<-0.35) \\ \\ P\left(x>9.8\right)=1-0.3632 \\ \\ P\left(x>9.8\right)=0.6368 \end{gathered}[/tex]
e.
No, you don't need to make the assumption
f.
Q1 = 0.25
In this case the value of z = 0.25, so we look for the closest value in the normal table, like this:
Thanks to this, we make the following equation:
[tex]\begin{gathered} -0.67=\frac{x-9.9}{\frac{1.8}{\sqrt{35}}} \\ \\ x-9.9=-0.19311 \\ \\ x=-0.1931+9.9 \\ \\ x=9.7069 \\ \\ Q_1=9.7069 \end{gathered}[/tex]
Q3 = 0.75
In this case the value of z = 0.75, so we look for the closest value in the normal table, like this:
Therefore:
[tex]\begin{gathered} -0.67=\frac{x-9.9}{\frac{1.8}{\sqrt{39}}} \\ \\ x-9.9=0.1931 \\ \\ x=0.1931+9.9 \\ \\ x=10.0931 \\ \\ Q_3=10.0931-9.7069 \end{gathered}[/tex]
Therefore, the interquartile range would be:
[tex]\begin{gathered} IQR=Q_3-Q_1 \\ \\ IQR=10.0931-9.7069 \\ \\ IQR=0.3862 \end{gathered}[/tex]
