Answered

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Suppose family incomes in a town are normally distributed with a mean of P25000 and a standard deviation of P6000 per month. What is the probability that a family has an income between P14,000 and P32,500?

Sagot :

The formula for the Z-score is,

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

Given:

[tex]\begin{gathered} x_1=P14,000,x_2=P32,500 \\ \mu=P25,000,\sigma=P6000 \end{gathered}[/tex]

Therefore,

[tex]\begin{gathered} Z_1=\frac{14000-25000}{6000}=-1.83333 \\ Z_2=\frac{32500-25000}{6000}=1.25 \end{gathered}[/tex]

Hence, the probability will be

[tex]P(Z_1Therefore, the answer is[tex]\begin{equation*} 0.86097 \end{equation*}[/tex]

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