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Consider a normal distribution of values with a mean of 32 and a standard deviation of 1.5. Find the probability that a value is less than 36.8.99.93%99.87%98.43%93.73%

Respuesta :

Finding Z:

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

Replacing the values given:

[tex]Z=\frac{36.8-32}{1.5}=3.2[/tex]

Then...

[tex]P(X<36.8)=P(Z<3.2)[/tex]

Using the Standard Normal Table, we can see that:

[tex]P(Z<3.2)=0.9993[/tex]

Answer: 99.93%

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