The probability of having L greater than Shelia's mean glucose level is 0.03 for L = 144.48.
Sheila's glucose level is shown to have a normal distribution, with a mean of 128 mg/dl and a standard deviation of 8 mg/dl.
L represents the patient's blood test level.
The probability that the mean glucose level, 128 mg/dl, is greater than L is given as 0.03.
To find L,
P(128>L) = 0.03
= 1 - P(128 ≤ L) = 0.03
=P(128 ≤ L) = 0.98
We know that we can find the probability of any normal distribution by converting it to standard form and then checking the p-value.
P(X < x) = Φ[(x - μ)/σ]
μ = mean
σ = standard deviation
=Φ[(L - 128)/8] = 0.98
Therefore ,
(L - 128)/8 = 2.06
= L - 128 = 16.48
= L = 128 + 16.48
= 144.48
= L
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