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C. Neilsen reported that children between the ages of 2 and 5 watch an average of 25 hours of television per week. Assume the variable is normally distributed and the standard deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomly selected, find the probability that the mean of the number of hours they watch television will be greater than 26.3 hours.

Please explain!

Sagot :

The following information about the mean and standard deviation has been provided:

ц=25, б=3,n=20

We need to compute Pr(X⁻≥26.3).

The corresponding z-value needed to be computed is:

[tex]Z=\frac{X-u}{o/\sqrt{n} } =\frac{26.3-25}{3/\sqrt{20} }=1.9379[/tex]

Therefor, we get that

[tex]Pr(X\geq 26.3)=Pr(Z\geq \frac{26.3-25}{3/\sqrt{20} } )=Pr(Z\geq 1.9379)\\=1-0.9737=0.026[/tex]

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