Answer:
B = 8.9 mg/dL
Explanation:
From the given question;
Mean = 9.4 mg/dL
Standard deviation = 0.04 mg/dL
P(X < A) = P(Z < (A - mean)/standard deviation)
Let B be the borderline of low calcium level and those considered not low.
P(X < B) = 0.10
P(Z < (B - 9.4)/0.4) = 0.10
Taking a Z value that corresponds to 0.10 from standard normal distribution table
(B - 9.4)/0.4 = -1.28
B - 9.4 = -1.28 * 0.4
B - 9.4 = -0 .512
B = -0.512 + 9.4
B = 8.888
B = 8.9 mg/dL (1 decimal place)
Answer:
Question not complete. The completed one is given below with solution
Explanation:
Calcium levels in people are normally distributed with a mean of 9.4 mg/dL and a standard deviation of 0.4mg/dL. Individuals with calcium levels in the bottom 10% of the population are considered to have low calcium levels. Find the calcium level that is the borderline between low calcium levels and those not considered low. Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.
Mean = 9.4 mg/dL
Standard deviation = 0.04 mg/dL
P(X < A) = P(Z < (A - mean)/standard deviation)
Let B be the borderline of low calcium level and those considered not low.
P(X < B) = 0.10
P(Z < (B - 9.4)/0.4) = 0.10
Take Z value corresponding to 0.10 from standard normal distribution table
(B - 9.4)/0.4 = -1.28
B = 8.89 mg/dL
