Answer in Statistics and Probability for Jado #248040
October 24th, 2022
We are given,
“mu=7”
“sigma=2”
To find these probabilities, the values are standardized and the probabilities read from the standard normal tables as below.
a.
“p(hgt9)=p(((h-mu)/sigma)gt(9-mu)/sigma)=p(Zgt(9-7)/2)=p(Zgt1)”
This can also be written as,
“1-p(Zlt1)=1-0.8413=0.1587”
Therefore “p(hgt9)=0.1587”
b.
“p(hlt6)=p(((h-mu)/sigma)lt(6-mu)/sigma)=p(Zlt(6-7)/2)=p(Zlt-0.5)”
From the standard normal tables,
“p(Zlt-0.5)=0.3085”
Hence, “p(hlt6)=0.3085”
c.
“p(5lt hlt8)”. On standardizing this we have,
“p((5-7)/2lt Zlt(8-7)/2)=p(-1lt Zlt 0.5)”
This probability can be written as,
“p(Zlt0.5)-p(Zlt-1)=0.6915-0.1587=0.5328”
Therefore, “p(5lt hlt 8)=0.5328” .
