The measurement X to be taken on an individual from a population has a Normal
distribution with mean u and standard deviation o. What proportion of the
individuals in the population that have their X measurements between u-20 and
u+o?
0.6826894809
0.954499876
0 0.8185946784
0 0.9973000656
O none of these

The measurement X to be taken on an individual from a population has a Normal distribution with mean u and standard deviation o What proportion of the individua class=

Respuesta :

Using the normal distribution, it is found that the correct option is 0.8185946784.

--------------------------

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula, which  in a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], for a measure X, is given by:

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

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, which is the percentile of measure X.

The proportion of measures between [tex]\mu - 2\sigma[/tex] and [tex]\mu + \sigma[/tex] is the p-value of Z when [tex]X = \mu + \sigma[/tex] subtracted by the p-value of Z when [tex]X = \mu - 2\sigma[/tex]. Thus:

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

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

[tex]Z = \frac{\sigma}{\sigma}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a p-value of 0.8413.

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

[tex]Z = \frac{\mu - 2\sigma - \mu}{\sigma}[/tex]

[tex]Z = \frac{-2\sigma}{\sigma}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a p-value of 0.0228.

0.8413 - 0.0228 = 0.8185.

Thus, the correct option is 0.8185946784.

A similar problem is given at https://brainly.com/question/24841527