Question

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes approximately 1/2 second to perform one full spin, with a standard deviation of 1/3 second. What is
the approximate probability that it will take this child over 55 seconds to complete spinning?

Answer 1

Answer:

$P(X \geq0.55) \leq 0.22$

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

$Y=X_1+X_2+...+X_{100}$ is approximately normally distributed with

Y ~ N (100 × $\frac{1}{3^2}$ ) = N ( 50, $\frac{100}{9}$ )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, $\frac{100}{9}$ )

$P(Y>55) =P(Z>\frac{55-50}{10/3})$

$P(Y>55) =P(Z>1.5)$

$P(Y>55) =\phi (-1.5)$

$P(Y>55) =0.0668$

Using Chebyshev's inequality:

$P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}$

Let assume that X has a symmetric distribution:

Then:

$2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}$

$2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}$

$2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}$               where: ($\sigma^2 = \frac{1}{3^2/100}$)

$P(X \geq0.55) \leq 0.22$