Answer :
Final answer:
We used the z-score formula to calculate the percentage of kids who played between 4.8 and 5.5 hours outdoors per week. This percentage was subtracted from the 50% of kids who played less than the mean number of hours. Approximately 2096 children spent between 4.8 and 5.5 hours outdoors weekly.
Explanation:
In this problem, we're dealing with a normal distribution where your mean (μ) is equal to 4.8 hours, and your standard deviation (σ) is equal to 0.5 hours. Our task is to thus find the number of children who played between 4.8 and 5.5 hours outdoors per week.
We start by calculating the z-score for 5.5 hours using the formula Z = (X - μ) / σ. Where X is the value we are interested in (5.5 hours in this case).
Using this formula, Z = (5.5 - 4.8) / 0.5 = 1.4
A z-score of 1.4 corresponds to approximately 91.92% of the data in a normal distribution (this value is obtained from a z-table or by using appropriate software).
Since the mean (4.8 hours) is also the median in a normal distribution, we know that 50% of the data is below this. Hence the percentage of children playing between 4.8 and 5.5 hours is 91.92% - 50% which equals 41.92%.
To find the amount of children, we calculate 41.92% of 5000 kids which gives us roughly 2096 children spent between 4.8 hours and 5.5 hours outdoors per week.
Learn more about Normal Distribution here:
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