In the following problems, assume that x has a normal distribution with the
specified mean and standard deviation. Find the indicated probabilities. a. P(8≤x≤ 12); p = 15; o = 3.2
b. P(x ≥ 30); μ = 20; o = 3.4
a. P(8 ≤x≤ 12); p = 15; o = 3.2
b. P(x ≥30); μ = 20; o = 3

Respuesta :

Using the normal distribution, it is found that the indicated probabilities are given as follows:

a) P(8≤x≤ 12) = 0.1593 = 15.93%.

b. P(x ≥30) = 0.0004 = 0.04%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

Item a:

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 15, \sigma = 3.2[/tex].

The probability is the p-value of Z when X = 12 subtracted by the p-value of Z when X = 8, hence:

X = 12:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{12 - 15}{3.2}[/tex]

Z = -0.94

Z = -0.94 has a p-value of 0.1736.

X = 8:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{8 - 15}{3.2}[/tex]

Z = -2.19

Z = -2.19 has a p-value of 0.0143.

0.1736 - 0.0143 = 0.1593.

Item b:

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 20, \sigma = 3[/tex]

The probability is one subtracted by the p-value of Z when X = 30, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{30 - 20}{3}[/tex]

Z = 3.33

Z = 3.33 has a p-value of 0.9996.

1 - 0.9996 = 0.0004.

More can be learned about the normal distribution at https://brainly.com/question/27893812

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