Answer:
(a) 0.119
(b) 0.1699
Solution:
As per the question:
Mean of the emission, 
million ponds/day
Standard deviation, 
million ponds/day
Now,
(a) The probability for the water pollution to be at least 15 million pounds/day:
= 1 - P(Z < 1.178)
Using the Z score table:
= 1 - 0.881 = 0.119
The required probability is 0.119
(b) The probability when the water pollution is in between 6.2 and 9.3 million pounds/day:
P(Z < - 0.86) - P(Z < - 1.96)
Now, using teh Z score table:
0.1949 - 0.025 = 0.1699
1/9 because i you plug in two of the points into y2 -y1 over x2 -x1 thats what you get.
X=0.84....................
40 goes into 32 0 times 40 goes into 321 8 times 40 times 8 equls 320 so 8r1
Answer:
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 14 - 1 = 13
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 3.0123
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 664.14 - 224.85 = $439.29
The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
