Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm
al distribution with mean μ=6 ml and standard deviation σ=0.2 ml is a reasonable model for the distribution of the variable x = amount of red dye in the paint mixture. Use the normal distribution model to calculate the following probabilities. (Round all answers to four decimal places.)
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
(a) P(x > 6) =
This is 1 subtracted by the pvalue of Z when X = 6. So
has a pvalue of 0.5.
1 - 0.5 = 0.5.
(b) P(x < 6.2)=
This is the pvalue of Z when X = 6.2. So
has a pvalue of 0.8413
(c) P(x ≤ 6.2) =
In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.
(d) P(5.8 < x < 6.2) =
This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.
X = 6.2
has a pvalue of 0.8413
X = 5.8
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
(e) P(x > 5.7) =
This is 1 subtracted by the pvalue of Z when X = 5.7.
has a pvalue of 0.0668
1 - 0.0668 = 0.9332
(f) P(x > 5) =
This is 1 subtracted by the pvalue of Z when X = 5.