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, 4…

Question

asked Nov 26, 2022 199k views

1 Answer

Unifreak · Answer 1 · 2022-11-30T13:38:01+0000

Answer:

P=8.869

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

$A=(1-(1)/(k^2))*100\%$

Where

$k=(\=x-\mu)/((\sigma)/(\sqrt n))}}$

$k=(43-38)/((52)/(\sqrt 16))}}$

k=2.77

Therefore

Probability

P=(1-(1)/(2.77^2))

P=8.869