Answer:
<em>The probability of spending between 4 and 7 days in recovery</em>
<em>P(4≤x≤7) = 0.5445</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given mean of the Population μ = 5.3 days
Given standard deviation of the population 'σ' = 2 days
Let 'X' be the random variable in normal distribution
Let x₁ = 4
![Z_{1} = \frac{x_{1}-mean }{S.D} = \frac{4-5.3}{2} = -0.65](https://tex.z-dn.net/?f=Z_%7B1%7D%20%3D%20%5Cfrac%7Bx_%7B1%7D-mean%20%7D%7BS.D%7D%20%3D%20%5Cfrac%7B4-5.3%7D%7B2%7D%20%3D%20-0.65)
Let x₂ = 7
![Z_{2} = \frac{x_{2}-mean }{S.D} = \frac{7-5.3}{2} = 0.85](https://tex.z-dn.net/?f=Z_%7B2%7D%20%3D%20%5Cfrac%7Bx_%7B2%7D-mean%20%7D%7BS.D%7D%20%3D%20%5Cfrac%7B7-5.3%7D%7B2%7D%20%3D%200.85)
<u><em>Step(ii):-</em></u>
<u><em>The probability of spending between 4 and 7 days in recovery</em></u>
<em>P(4≤x≤7) = P(-0.65≤Z≤0.85)</em>
<em> = P(Z≤0.85) - P(Z≤-0.65)</em>
<em> = 0.5 + A( 0.85) - ( 0.5 - A(-0.65) </em>
<em> = 0.5 + A( 0.85) - 0.5 +A(0.65) ( ∵A(-0.65) = A(0.65)</em>
<em> = A(0.85) + A(0.65)</em>
<em> = 0.3023 + 0.2422</em>
<em> = 0.5445</em>
<u><em>Final answer:</em></u><em>-</em>
<em>The probability of spending between 4 and 7 days in recovery</em>
<em>P(4≤x≤7) = 0.5445</em>
<em> </em>