Respuesta :
Answer:
[tex]P(1739<X<1800)=0.0223[/tex]
(Correct to 4 decimal places)
Step-by-step explanation:
The probability of a continuous normal variable X found in a particular interval [a, b] is the area under the curve bounded by x=a and x=b and is given by:
[tex]P(a<X<b)={\int_{{a}}^{{b}}} f{{\left({X}\right)}}{\left.{d}{x}\right}[/tex]
where
[tex]f(X)=\frac{1}{\sigma \sqrt{2 \pi} }e^{-\frac{1}{2} (\frac{x-u}{ \sigma} )^2}[/tex]
[tex]f(X)=\frac{1}{200 \sqrt{2 \pi} }e^{-\frac{1}{2} (\frac{x-1400}{ 200} )^2}\\=\frac{1}{200 \sqrt{2 \pi} }e^{- \frac{(x-1400)^2}{ 80000}\\[/tex]
[tex]P(1739<X<1800)={\int_{{1739}}^{{1800}}} f{{\left({X}\right)}}{\left.{d}{x}\right}[/tex]
[tex]P(1739<X<1800)={\int_{{1739}}^{{1800}}} \frac{1}{200 \sqrt{2 \pi} }e^{- \frac{(x-1400)^2}{ 80000}\\[/tex]
Using a calculator,
[tex]P(1739<X<1800)=0.0223[/tex]
Answer:
The probability that the weight of a randomly selected steer is between 1,739 and 1,800 lbs is 0.0218 or 2.18%
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
μ of the weights of steers in a herd= 1,400 lbs
σ² = 40,000 ⇒ σ = √40,000 = 200 lbs
2. Find the probability that the weight of a randomly selected steer is between 1,739 and 1,800 lbs. Round your answer to four decimal places.
Let's find the z-score for 1,739 and 1.800, this way:
z-score = (X - μ)/σ
z-score = (1,739 - 1,400)/200
z-score = 339/200
z-score = 1.695 = 1.70 (rounding to the next hundredth)
z-score = (X - μ)/σ
z-score = (1,800 - 1,400)/200
z-score = 400/200
z-score = 2
Now, let's find p for z-score = 1.7 and p for z-score = 2, using the z-table, as follows:
p (z = 1.7) = 0.9554
p (z = 2) = 0.9772
In consequence,
p (1.7 ≤ z ≤ 2 ) = 0.9772 - 0.9554
p (1.7 ≤ z ≤ 2 ) = 0.0218
The probability that the weight of a randomly selected steer is between 1,739 and 1,800 lbs is 0.0218 or 2.18%
