Respuesta :
Answer:
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
