contestada

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X= percent of fat calories.(a) Find the z-score corresponding to 30 percent of fat calories, rounded to 3 decimal places.

Respuesta :

The formula for Z-score is given as,

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

Given that

[tex]\begin{gathered} X=percent\text{ of fat calories=30} \\ \mu=\operatorname{mean}\text{ = 36} \\ \sigma=10 \end{gathered}[/tex]

Therefore,

[tex]\begin{gathered} Z=\frac{30-36}{10}=\frac{-6}{10}=-0.6 \\ \therefore Z=-0.6 \end{gathered}[/tex]

Hence,

[tex]\begin{gathered} Z=-0.6\approx-0.600(3\text{ decimal places)} \\ \therefore Z=-0.600 \end{gathered}[/tex]

Otras preguntas