Question

6-35 A recent study by an environmental watchdog determined that the amount of contaminants in Minnesota lakes (in parts per million) it has a normal distribution with a mean of 64 ppm and variance of 17.6. Assume that 35 lakes are randomly selected and sampled. Find the probability that the sample average of the amount of contaminants is a) Greater than 72 ppm. b) Between 64 and 72 ppm. c) Exactly 64 ppm. d) Greater than 94 ppm.

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μ = 64
σ = √(17.6/35) = 0.7091242083
a)
Z score =(x - μ)/σ
=(72 - 64)/0.7091242083
=11.28152
P-value from Z-Table:
P(x<72) = 1
P(x>72) = 1 - P(x<72) = 1 - 1
P(x>72) = 0
b)
P(64 < x < 72) = P(x < 72) - 0.5 = 0.5
P(64 < x < 72) = 0.5
c)
The probability of any single value from a continuous distribution is zero
P(64) = 0
d)
Z score =(x - μ)/σ
=(94 - 64)/0.7091242083
=42.30571
P-value from Z-Table:
P(x<94) = 1
P(x>94) = 1 - P(x<94) = 1 - 1
P(x>94) = 0

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