Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
Answer:
78
Step-by-step explanation:
7+8/2 multiply by 6
Answer:
given,
mp= 4500
now
discount amount=dis%of mp
=x%of 4500
=45x
now,
sp=mp- discount
=4500-45x
now,
vat=vat%of SP
=10%*4500-45x
=450-45x
now,
sp with vat=sp+vat amount
=4500-45x+450-45x
=4950-90x
now,
4950-90x=4400
or,-90x=-550
or, x=55/9
or, x=6.11%
therefore, dis percent is 6.11%
Answer: 80
Step-by-step explanation:
