At one point the average price of regular unleaded gasoline was $3.54 per gallon. assume that the standard deviation price per gallon is $0.06 per gallon and use chebyshev's inequality to answer the following.
(a) what percentage of gasoline stations had prices within 3 standard deviations of the mean?
(b) what percentage of gasoline stations had prices within 2.5 standard deviations of the mean? what are the gasoline prices that are within 2.5 standard deviations of the mean?
(c) what is the minimum percentage of gasoline stations that had prices between $3.30 and $3.78?

a) at least 88.88% of the gasoline stations had prices within 3 standard deviations of the mean

b) at least 84% of the gasoline stations had prices within 2.5 standard deviations of the mean

c) at least 93.75% of the gasoline stations had prices between $3.30 and $3.78 per gallon

Step-by-step explanation:

Following Chebyshev's inequality:

P(|X- μ | ≤ k*σ|) ≥ 1- 1/k²

where

X= gasoline price

μ = expected value of X (mean)

σ = standard deviation of X

P(|X- μ | ≤ k*σ|) = probability that the random variable X is within k standard deviations of the mean

a) since k=3

P(|X- μ | ≤ 3*σ|) ≥ 1- 1/3² = 8/9 = 88.88%

P(|X- μ | ≤ 3*σ|) ≥ 88.88%

therefore at least 88.88% of the gasoline stations had prices within 3 standard deviations of the mean

b) for k=2.5

P(|X- μ | ≤ 2.5*σ|) ≥ 1- 1/2.5² = 0.84 = 84%

P(|X- μ | ≤ 2.5*σ|) = 84%

therefore at least 84% of the gasoline stations had prices within 2.5 standard deviations of the mean

c) starting from

|X- μ | = k*σ

k= |X- μ | / σ

for X=$3.30 , μ=$3.54, σ= $0.06

k= |X- μ | / σ = |3.30 - 3.54| / 0.06 = 4

for X=$3.78 , μ=$3.54, σ= $0.06

k= |X- μ | / σ = |3.78- 3.54| / 0.06 = 4

then, since the range 3.78 - 3.30 is the same that 4 standard deviations of the mean, we apply  Chebyshev's inequality for k=4 :

P( 3.30 ≤X≤3.78 ) = P(|X- μ | ≤ 4*σ|) ≥ 1- 1/4² = 0.9375 = 93.75%

P( $3.30 ≤X≤$3.78 ) ≥ 93.75%

thus at least 93.75% of the gasoline stations had prices between $3.30 and $3.78 per gallon