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

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

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 ×
`(1)/(3^2)` ) = N ( 50,
`(100)/(9)` )

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

N ( 50,
`(100)/(9)` )

`P(Y>55) =P(Z>(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 (\sigma^2)/(K^2)`

Let assume that X has a symmetric distribution:

Then:

`2P(X-\mu\geq K)\leq) (\sigma^2)/(K^2)`

`2P(X \geq \mu+K)\leq) (\sigma^2)/(K^2)`

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

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