Question

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)

Answer 1

Answer:

$P=8.869$

Step-by-step explanation:

From the question we are told that:

Mean $\=x =38$

Variance $\sigma=52$

Sample size $n=16$

$X=(33, 43)$

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

$A=(1-\frac{1}{k^2})*100\%$

Where

$k=\frac{\=x-\mu}{\frac{\sigma}{\sqrt n}}}}$

$k=\frac{43-38}{\frac{52}{\sqrt 16}}}}$

$k=2.77$

Therefore

Probability

$P=(1-\frac{1}{2.77^2})$

$P=8.869$