Respuesta :
Answer:
($3.61, $4.27) is the range in which at least 88.9% of the data will reside.
Step-by-step explanation:
Chebyshev's Theorem
- This theorem states that at least [tex]1 - \dfrac{1}{k^2}[/tex] percentage of data lies within k standard deviation from the mean.
- For k = 2
[tex]\mu \pm 2\sigma\\\\1 - \dfrac{1}{(2)^2}\% = 75\%[/tex]
Atleast 75% of data lies within two standard deviation of the mean for a non-normal data.
- For k = 3
[tex]\mu \pm 3\sigma\\\\1 - \dfrac{1}{(3)^2}\% = 88.9\%[/tex]
Thus, range of data for which 88.9% of data will reside is
[tex]\mu \pm 3\sigma\\= (\mu - 3\sigma , \mu + \3\sigma)[/tex]
Now,
Mean = $3.94
Standard Deviation = $0.11
Putting the values, we get,
Range =
[tex]3.94 \pm 30.11\\= (3.94 - 3(0.11) , 3.94 + 3(0.11))\\=(3.61,4.27)[/tex]
($3.61, $4.27) is the range in which at least 88.9% of the data will reside.
Answer:
Using Chebyshev's Theorem, the range in which at least 88.9% of the data reside is [$3.82, $4.06].
Step-by-step explanation:
Given:
The mean of the prices of a gallon of milk is, [tex]\mu=\$3.94[/tex]
The standard deviation of the prices of a gallon of milk is, [tex]\sigma=\$0.11[/tex]
The Chebyshev's Theorem states that for a random variable X with finite mean μ and finite standard deviation σ and for a positive constant k the provided inequality exists,
[tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}[/tex]
The value of [tex]\frac{\sigma^{2}}{k^{2}} = 0.889[/tex]
Then solve for k as follows:
[tex]\frac{\sigma^{2}}{k^{2}} = 0.889\\\frac{(0.11)^{2}}{k^{2}}=0.889\\k^{2}=\frac{(0.11)^{2}}{0.889}\\k=\sqrt{0.0136108} \\\approx0.1167[/tex]
The range in which at least 88.9% of the data will reside is:
[tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}\\P(\mu-k\leq X\leq \mu+k)\leq 0.889\\P(3.94-0.1167\leq \leq X\leq 3.94+0.1167)\leq 0.889\\P(3.8233\leq X\leq 4.0567)\leq 0.889\\\approxP(3.82\leq X\leq 4.06)\leq 0.889[/tex]
Thus, the probability of prices of a gallon of milk between $3.82 and $4.06 is 0.889.
