Suppose a variable has a normal distribution with mean 67 and standard deviation 4. What percentage of the distribution is less than 75? (Use z-score.)

Respuesta :

Answer:

The percentage of the distribution is less than 75 is 97.72%.

Step-by-step explanation:

Given,

Mean of the distribution,

[tex]\mu=67[/tex]

Standard deviation,

[tex]\sigma = 4[/tex]

Thus, the z-score of the score 75,

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

[tex]=\frac{75-67}{4}[/tex]

[tex]=\frac{8}{4}[/tex]

[tex]=2[/tex]

With the help of z-score table,

[tex]P(x<75)=0.9772=97.72\%[/tex]

Hence, the percentage of the distribution is less than 75 is 97.72%.

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