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(h\\gt9)=p(((h-\\mu)\/\\sigma)\\gt(9-\\mu)\/\\sigma)=p(Z\\gt(9-7)\/2)=p(Z\\gt1)"
This can also be written as,
"1-p(Z\\lt1)=1-0.8413=0.1587"
Therefore "p(h\\gt9)=0.1587"
b.
"p(h\\lt6)=p(((h-\\mu)\/\\sigma)\\lt(6-\\mu)\/\\sigma)=p(Z\\lt(6-7)\/2)=p(Z\\lt-0.5)"
From the standard normal tables,
"p(Z\\lt-0.5)=0.3085"
Hence, "p(h\\lt6)=0.3085"
c.
"p(5\\lt h\\lt8)". On standardizing this we have,
"p((5-7)\/2\\lt Z\\lt(8-7)\/2)=p(-1\\lt Z\\lt 0.5)"
This probability can be written as,
"p(Z\\lt0.5)-p(Z\\lt-1)=0.6915-0.1587=0.5328"
Therefore, "p(5\\lt h\\lt 8)=0.5328" .
Comments
Leave a comment