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ella [17]
3 years ago
8

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a

pproximately 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?
Mathematics
1 answer:
Sati [7]3 years ago
6 0

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

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