Question

1) What is the probability of a randomly selected value from a normally distributed population falling within 1.5 standard deviations of the mean?

8) What is the probability of a randomly selected value from a normally distributed population NOT being between 0.68 standard deviations below the mean and 1.5 standard deviations above the mean?

***For the following questions, assume a business has an average daily revenue of $1200 and revenue levels are found to be normally distributed with a standard deviation of $120.

11) What is the probability revenue on a randomly selected day is greater than $1300?

15) What is the probability revenue on a randomly selected day is not between $1050 and $1300?

20) 70% of all daily revenue values within how many dollars of the mean?

Answer 1

Solution(1)
We need to calculate
P(-1.5<Z<1.5) = P(Z<1.5) - P(Z<-1.5)
From Z table we found p-values
P(-1.5<Z<1.5) = P(Z<1.5) - P(Z<-1.5) = 0.9332 - 0.0668 = 0.8664
Solution(8)
P(0.68>Z>1.5) = P(Z<0.68) + P(Z>1.5) = 0.7518 + 0.0668 = 0.8186
Solution(11)
Mean = 1200, Standard deviation = 120
P(X>1300) = 1- P(X<1300)
Z = (1300-1200)/120 = 0.83
P(X>1300) = 1 - 0.7967 = 0.2033
Solution(15)
P(1050>X>1300) = P(X<1050) + P(X>1300)
Z = (1050-1200)/120 = -1.25
Z = (1300-1200)/120 = 0.83
P(1050>X>1300) = P(X<1050) + P(X>1300) = 0.1065 + 0.2033 = 0.3098
Solution(20)
p-values are 0.15 and 0.85 so Z scores are -1.036 and 1.036
-1.036 = (X-1200)/120
X = 1075.68
1.036 = (X-1200)/120
X = 1324.32
P(1075.68<X<1324.32) = 0.7