The age of children in kindergarten on the first day of school is uniformly distributed between 4.91 and 5.97 years old. A first time kindergarten child is selected at random. Round answers to 4 decimal places if possible. a. The mean of this distribution is b. The standard deviation is c. The probability that the the child will be older than 5 years old?

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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**Problem Statement:**

The age of children in kindergarten on the first day of school is uniformly distributed between 4.91 and 5.97 years old. A first-time kindergarten child is selected at random. Round answers to 4 decimal places if possible.

**Questions:**

a. The mean of this distribution is ______.

b. The standard deviation is ______.

c. The probability that the child will be older than 5 years old ______.

**Explanation:**

In this problem, we are dealing with a uniform distribution. The ages of the children are distributed evenly between 4.91 and 5.97 years. The mean and standard deviation are calculated using the properties of uniform distributions.

- **Mean:** For a uniform distribution, the mean is calculated as \((a + b) / 2\), where \(a\) and \(b\) are the lower and upper bounds, respectively.

- **Standard Deviation:** For a uniform distribution, the standard deviation is calculated as \(\sqrt{((b - a)^2) / 12}\).

- **Probability (c):** To find the probability that a child is older than 5 years, we calculate the relative position of 5 within the distribution range.
Transcribed Image Text:**Problem Statement:** The age of children in kindergarten on the first day of school is uniformly distributed between 4.91 and 5.97 years old. A first-time kindergarten child is selected at random. Round answers to 4 decimal places if possible. **Questions:** a. The mean of this distribution is ______. b. The standard deviation is ______. c. The probability that the child will be older than 5 years old ______. **Explanation:** In this problem, we are dealing with a uniform distribution. The ages of the children are distributed evenly between 4.91 and 5.97 years. The mean and standard deviation are calculated using the properties of uniform distributions. - **Mean:** For a uniform distribution, the mean is calculated as \((a + b) / 2\), where \(a\) and \(b\) are the lower and upper bounds, respectively. - **Standard Deviation:** For a uniform distribution, the standard deviation is calculated as \(\sqrt{((b - a)^2) / 12}\). - **Probability (c):** To find the probability that a child is older than 5 years, we calculate the relative position of 5 within the distribution range.
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