Respuesta :

Answer:

[tex]0.30854[/tex] or [tex]30.854\%[/tex].

Step-by-step explanation:

We have been given that a normal distribution has μ=80 and σ=10. We are asked to find the the probability of randomly selecting a score greater than 85 from this distribution.

First of all, we will find the z-score corresponding to normal score 85 as:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Upon substituting our given values in z-score formula, we will get:

[tex]z=\frac{85-80}{10}[/tex]

[tex]z=\frac{5}{10}[/tex]

[tex]z=0.5[/tex]

Now, we need to find [tex]P(z>0.5)[/tex] for the probability of randomly selecting a score greater than 85 from this distribution.

Using formula [tex]P(z>a)=1- P(z<a)[/tex], we will get:

[tex]P(z>0.5)=1-P(z<0.5)[/tex]

[tex]P(z>0.5)=1-0.69146[/tex]

[tex]P(z>0.5)=0.30854[/tex]

Therefore, the probability of randomly selecting a score greater than 85 from this distribution would be 0.30854 or [tex]30.854\%[/tex].

Using the normal distribution, it is found that there is a 0.3085 = 30.85% probability of randomly selecting a score greater than 85 from this distribution.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

  • It 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 X.

In this problem:

  • The mean is of [tex]\mu = 80[/tex].
  • The standard deviation is of [tex]\sigma = 10[/tex].

The probability of randomly selecting a score greater than 85 from this distribution is 1 subtracted by the p-value of Z when X = 85, hence:

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

[tex]Z = \frac{85 - 80}{10}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a p-value of 0.6915.

1 - 0.6915 = 0.3085.

0.3085 = 30.85% probability of randomly selecting a score greater than 85 from this distribution.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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