g From a distribution with mean 38 and variance 52, a sample of size 16 is taken. Let X be the mean of the sample. Show that the probability is at least 0.87 that X is in (33, 43)

Respuesta :

Answer:

[tex]P=8.869[/tex]

Step-by-step explanation:

From the question we are told that:

Mean [tex]\=x =38[/tex]

Variance [tex]\sigma=52[/tex]

Sample size [tex]n=16[/tex]

[tex]X=(33, 43)[/tex]

Generally the equation for Chebyshev's Rule  is mathematically given by

[tex]A=(1-\frac{1}{k^2})*100\%[/tex]

Where

[tex]k=\frac{\=x-\mu}{\frac{\sigma}{\sqrt n}}}}[/tex]

[tex]k=\frac{43-38}{\frac{52}{\sqrt 16}}}}[/tex]

[tex]k=2.77[/tex]

Therefore

Probability

[tex]P=(1-\frac{1}{2.77^2})[/tex]

[tex]P=8.869[/tex]