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Mrs. Burleson surveyed the students in her class to find the number of minutes they spent doing homework each night. she found that the data was normally distributed with U=30 and O=10.How much time spent on homework would you expect from 2.5% of students with the least time spent on homework?

Sagot :

With the help of the 68-95-99 rule (see picture above) we know that those values with a z-score less than -2 are located in the 2.5% zone on the left.

Z-score is computed as follows:

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

where x is the observed value, μ is the mean and σ is the standard deviation.

Substituting with z = -2, μ = 30, and σ = 10, we get:

[tex]\begin{gathered} -2=\frac{x-30}{10} \\ (-2)\cdot10=x-30 \\ -20+30=x \\ 10=x \end{gathered}[/tex]

It is expected that 2.5% of the students spend less than 10 minutes.

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