Question

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal 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.)

(a) P(x > 6) =

(b) P(x < 6.2)=

(c) P(x ≤ 6.2) =

(d) P(5.8 < x < 6.2) =

(e) P(x > 5.7) =

(f) P(x > 5) =

Answer 1

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 6, \sigma = 0.2`

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

`Z = (X - \mu)/(\sigma)`

`Z = (6-6)/(0.2)`

`Z = 0`

`Z = 0` 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

`Z = (X - \mu)/(\sigma)`

`Z = (6.2-6)/(0.2)`

`Z = 1`

`Z = 1` 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

`Z = (X - \mu)/(\sigma)`

`Z = (6.2-6)/(0.2)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

X = 5.8

`Z = (X - \mu)/(\sigma)`

`Z = (5.8-6)/(0.2)`

`Z = -1`

`Z = -1` 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.

`Z = (X - \mu)/(\sigma)`

`Z = (5.8-6)/(0.2)`

`Z = -1.5`

`Z = -1.5` 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.

`Z = (X - \mu)/(\sigma)`

`Z = (5-6)/(0.2)`

`Z = -5`

`Z = -5` has a pvalue of 0.

1 - 0 = 1