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Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.8 hours and the amount of time spent alone is normally distributed. We randomly survey one 4-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. What percent of the children spend over 10 hours per day unsupervised?

Respuesta :

A Z-score helps us to understand how far is the data from the mean. The percentage of children who spend over 10 hours per day unsupervised is 0.005%.

What is Z-score?

A Z-score helps us to understand how far is the data from the mean. It is a measure of how many times the data is above or below the mean. It is given by the formula,

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

Where Z is the Z-score,

X is the data point,

μ is the mean and σ is the standard variable.

Given that the average time that is unsupervised is 3 hours and the standard deviation is 1.8 hours. The percentage of children who spend over 10 hours per day unsupervised is,

P(X>10) = 1 - P(X<10)

            [tex]= 1 - P(Z < \dfrac{10-3}{1.8})[/tex]

            = 1 - P(Z< 3.889)

            = 1 - 0.99995

            = 0.00005

Hence, The percentage of children who spend over 10 hours per day unsupervised is 0.005%.

Learn more about Z-score:

https://brainly.com/question/13299273

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