Answer 1:
Looking at the provided data, we can see that the closest z-value to 1.153 is 1.132.
Therefore, the coordinates for the point 1.153 would be (1.132, 0.878).
Answer 2:
S = sigma/sqrt(n)
S = 1.6/sqrt(n)
S = 0.32
Now
Z = x-u/s
Z = 70.5-70/0.32
Z = 1.56
By using normal table we got:
P(70.5 ≤ x ) = 0.9409
Answer 3:
a=(μ/σ)2−4.0
a=(41.5692/20.7846)2−4.0
a≈(1.9996)2−4.0
a≈4.0−4.0
a=0
Now, we can standardize the variable �X using the formula:
Z=X−μ/σ
Z=31.2-41.5692/20.7846
Z≈−0.4998
P(X>31.2)=1−F(X≤31.2;a)
==1−(1−e−λx)
=e−λx
≈e−0.7516
Using a calculator, we find:
e−0.7516≈0.4726
P(X>31.2) is 0.4726