The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.
Find the z score for 93.8 inches:
93.8 inches - 91.0 inches 2.8 inches
z = ------------------------------------- = ----------------- = 1.4
2 inches 2 inches
Now find the area under the standard normal curve to the left of z = +1.4.
My calculator returns the following:
normalcdf(-100,1.4) = 0.919. This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.
One way is to estimate
1^2=1
2^2=4
3^2=9
4^2=16
5^2=25
6^2=36
7^2=49
8^2=64
the number 61 is between 49 and 64 therefor the square root of 61is between the intigers 7 and 8 or -7 and -8
Answer:
1=1 times 2/2 1 + 2=2 times 3/2 1 +2+3=3 times 4/2 1+2+3+4=4 times 5/2 1+2+3+4+5=5 times 6/2
<span>150%
</span>Get 150 and multiply it by (34/100).
<span>In other words, do this: 150 x 0.34 (The 0.34 represents 34%. 0.76 would be 76%, etc) </span>
<span>This gives you: 51 </span>
