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If z is a standard normal variable, find the probability. The Probability that z is greater than -1.82.

A. -0.0344
B. 0.9656
C. 0.0344
D. 0.4656

Respuesta :

SonOfZeus

Answer:

C.) 0.9656

Step-by-step explanation:

Using a z-table look for -1.82 and we obtain : 0.0344.  So what's the probability that z > -1.82 is the same as saying when is z > 0.0344? Since we're talking about probabilities of an outcome from ranging from 0% to 100% meaning 0 through 1 in decimals then:

1 - 0.0344 = 0.9656

So the probability that z>-1.82 is C.) 0.9656

Using the normal distribution, it is found that there is a 0.9656 probability that z is greater than -1.82, option B.

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.
  • The probability of finding a value above X, that is, having a z-score above the one found, is 1 subtracted by the p-value of Z.

In this problem:

z = -1.82 has a p-value of 0.0344.

1 - 0.0344 = 0.9656.

0.9656 probability that z is greater than -1.82, option B.

A similar problem is given at https://brainly.com/question/5954134

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