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: B
Step-by-step explanation:
i took the quiz and got it right
Answer:
26 - 7 i. hopefully this helps!
Answer: Mr Drysdale deposited 14,500
Step-by-step explanation:
Using the simple Interest formula
Simple Interest = Principal(Initial Money) * Interest Rate * Time Period
I = P*R*T
I = PRT
To get P which is the initial money he deposited, we divide both sides by RT
I/RT = PRT/RT
I/RT = P
Therefore the money deposited
P = I/RT
Interest (I)= 906.25
Interest Rate (R)= 6.25% = { converting percentage to decimal} (6.25/100)= 0.0625
Time Period (T) = 1 year
Principal (P)= ?
P= I/RT
P= 906.25/(0.0625 * 1)
P= 906.25/0.0625
P= 1450
The money he deposited in his local bank is 14,500
