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
Answer:
130.8
Step-by-step explanation:
i just think that you times 0.40 and 327 and then press = and it should give you 130.8
i hope it works
Answer:
(C) y= 2.11x
Step-by-step explanation:
Answer:
1.16
Step-by-step explanation:
Given that;
For some positive value of Z, the probability that a standardized normal variable is between 0 and Z is 0.3770.
This implies that:
P(0<Z<z) = 0.3770
P(Z < z)-P(Z < 0) = 0.3770
P(Z < z) = 0.3770 + P(Z < 0)
From the standard normal tables , P(Z < 0) =0.5
P(Z < z) = 0.3770 + 0.5
P(Z < z) = 0.877
SO to determine the value of z for which it is equal to 0.877, we look at the
table of standard normal distribution and locate the probability value of 0.8770. we advance to the left until the first column is reached, we see that the value was 1.1. similarly, we did the same in the upward direction until the top row is reached, the value was 0.06. The intersection of the row and column values gives the area to the two tail of z. (i.e 1.1 + 0.06 =1.16)
therefore, P(Z ≤ 1.16 ) = 0.877
