Question

The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2 days. What is the probability of spending between 4 and 7 days in recovery? (Round your answer to four decimal places.)

Answer 1

Answer:

The probability of spending between 4 and 7 days in recovery

P(4≤x≤7) = 0.5445

Step-by-step explanation:

Step(i):-

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

$Z_{1} = \frac{x_{1}-mean }{S.D} = \frac{4-5.3}{2} = -0.65$

Let    x₂ = 7

$Z_{2} = \frac{x_{2}-mean }{S.D} = \frac{7-5.3}{2} = 0.85$

Step(ii):-

The probability of spending between 4 and 7 days in recovery

P(4≤x≤7) = P(-0.65≤Z≤0.85)

             =  P(Z≤0.85) - P(Z≤-0.65)

            = 0.5 + A( 0.85) - ( 0.5 - A(-0.65)

            = 0.5 + A( 0.85) -  0.5 +A(0.65)   ( ∵A(-0.65) = A(0.65)

           =   A(0.85) + A(0.65)

          = 0.3023 + 0.2422

         = 0.5445

Final answer:-

The probability of spending between 4 and 7 days in recovery

P(4≤x≤7) = 0.5445