Questions: Suppose that the age of students at George Washington Elementary school is uniformly distributed between 5 and 11 years old. 37 randomly selected children from the school are asked their age. Round all answers to 4 decimal places where possible. a. What is the distribution of X ? X-U(5 , 11 ) Suppose that 37 children from the school are surveyed. Then the sampling distribution is b. What is the distribution of x̄ ? x̄-N 8.0 , 3/37 c. What is the probability that the average of 37 children will be between 7 and 7.5 years old? 0.0393

Suppose that the age of students at George Washington Elementary school is uniformly distributed between 5 and 11 years old. 37 randomly selected children from the school are asked their age. Round all answers to 4 decimal places where possible.
a. What is the distribution of X ? X-U(5 , 11 )
Suppose that 37 children from the school are surveyed. Then the sampling distribution is
b. What is the distribution of x̄ ? x̄-N 8.0 , 3/37
c. What is the probability that the average of 37 children will be between 7 and 7.5 years old? 0.0393

Transcript text: Suppose that the age of students at George Washington Elementary school is uniformly distributed between 5 and 11 years old. 37 randomly selected children from the school are asked their age. Round all answers to 4 decimal places where possible. a. What is the distribution of $X$ ? $X-U(5$ , 11 ) Suppose that 37 children from the school are surveyed. Then the sampling distribution is b. What is the distribution of $\bar{x}$ ? $\bar{x}-\mathrm{N} 8.0$ , 3/37 c. What is the probability that the average of 37 children will be between 7 and 7.5 years old? 0.0393

Solution

What is the distribution of $ X $?

Distribution of $ X $

The age of students at George Washington Elementary School is uniformly distributed between 5 and 11 years old. Therefore, we can express this as: \[ X \sim U(5, 11) \]

$\boxed{X \sim U(5, 11)}$

What is the distribution of $ \bar{x} $?

Distribution of $ \bar{x} $

The mean of the uniform distribution is given by: \[ \mu = \frac{a + b}{2} = \frac{5 + 11}{2} = 8.0 \] The variance of the uniform distribution is: \[ \sigma^2 = \frac{(b - a)^2}{12} = \frac{(11 - 5)^2}{12} = \frac{36}{12} = 3 \] The variance of the sample mean $ \bar{x} $ is: \[ \sigma^2_{\bar{x}} = \frac{\sigma^2}{n} = \frac{3}{37} \] Thus, the distribution of $ \bar{x} $ can be expressed as: \[ \bar{x} \sim N\left(8.0, \frac{3}{37}\right) \]

$\boxed{\bar{x} \sim N\left(8.0, \frac{3}{37}\right)}$

What is the probability that the average of 37 children will be between 7 and 7.5 years old?

Calculating the probability

To find the probability that the average age $ \bar{x} $ is between 7 and 7.5, we first calculate the cumulative distribution function (CDF) values at these points: \[ P(7 < \bar{x} < 7.5) = P(\bar{x} \leq 7.5) - P(\bar{x} \leq 7) \] The calculated probability is: \[ P(7 < \bar{x} < 7.5) = 0.0393 \]

$\boxed{P(7 < \bar{x} < 7.5) = 0.0393}$

The distribution of $ X $ is $ \boxed{X \sim U(5, 11)} $.

The distribution of $ \bar{x} $ is $ \boxed{\bar{x} \sim N\left(8.0, \frac{3}{37}\right)} $.

The probability that the average age is between 7 and 7.5 years is $ \boxed{0.0393} $.

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