In a certain distribution the mean is 50 with a standard dev

In a certain distribution, the mean is 50 with a standard deviation of 5. Use Chebyshev\'s Theorem to tefl the probability that a number is less than 30 or more than 70. The probability a number is less than 30 or more than 70 is at most (Round to the nearest thousandth if necessary.)

Solution

Mean = 50

SD=5

Mean- 4sd = 50-4*5 =30

Mean+4sd = 50+4*5 =70

K=4

Form Chebyshev\'s Theorem, 1-1/k²   of the values falls between ( mean-k*sd, mean+k*sd)

1-1/4²   = 0.9375 of the values falls between ( 30, 70)

P( x <30 or x >70 = 1-0.9375 =0.0625

Answer = 0.0623 ( three decimals)