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

The probability of spending between 4 and 7 days in recovery

P(4≤x≤7) = 0.5445

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) = (x_(1)-mean )/(S.D) = (4-5.3)/(2) = -0.65`

Let x₂ = 7

`Z_(2) = (x_(2)-mean )/(S.D) = (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