Assume that serum potassium levels in a normal healthy population are normally d

Question

Assume that serum potassium levels in a normal healthy population are normally distributed, with a mean of 4.5 mEq/L and a standard deviation of 0.6 mEq/L.

(3 pts) What proportion of the population has serum potassium levels between 4.2 and 4.8 mEq/L?

(4 pts) If repeated samples of 36 were selected, what proportion of these samples would have means xbar between 4.2 and 4.8 mEq/L?

(3 pts) One sample of size 36 had sample mean xbar= 4.4. What is the 95% confidence interval for the mean serum potassium in the population? (Assume the standard deviation in the population is 0.6 mEq/L.)

Extra credit (2 points) A researcher would like to do a study that derives a confidence interval for the mean serum potassium (similar to that in question 1.c). However, the researcher wants a margin of error of 0.08 mEq/L. How many participants are needed?

Explanation / Answer

Assume that serum potassium levels in a normal healthy population are normally distributed, with a mean of 4.5 mEq/L and a standard deviation of 0.6 mEq/L.

(a) What proportion of the population has serum potassium levels between 4.2 and 4.8 mEq/L?

Let say X is the serum potassiom level so we have to calculate

Pr( 4.2 <= X <= 4.8) = Pr( X<= 4.8) - Pr( X<= 4.2)

Z - value for X = 4.8 is Z = (4.8 - 4.5)/ 0.6 = 0.5

and for X = 4.2 Z - value = -0.5

so Pr (4.2 <= X <= 4.8) = (0.5) - (-0.5) = 0.6915 - 0.3085 = 0.383

HEre = cumulative normal probabiluty functiion.

(b) If repeated samples of 36 were selected, what proportion of these samples would have means xbar between 4.2 and 4.8 mEq/L?

Here n= 36

so standard error of the mean = s/n = 0.6/36 = 0.6/6 = 0.1

so Pr ( 4.2 <= xbar<= 4.8 ; 4.5; 0.1)

so here Z = (4.8 - 4.5)/ 0.1 = 3 so Z = +-3

so Pr ( 4.2 <= xbar<= 4.8 ; 4.5; 0.1) =  (3) -  (-3) = 0.9987- 0.0013 = 0.9974

(c) One sample of size 36 had sample mean xbar= 4.4. What is the 95% confidence interval for the mean serum potassium in the population? (Assume the standard deviation in the population is 0.6 mEq/L.)

95% confidence interval for mean serum potassium population = xbar+- Z0.95 (s/n )

95% CI = 4.4 +- 1.96 * (0.6/36) = 4.4 +- 1.96 * 0.1 = (4.204, 4.596)

(d) A researcher would like to do a study that derives a confidence interval for the mean serum potassium (similar to that in question 1.c). However, the researcher wants a margin of error of 0.08 mEq/L. How many participants are needed?

margin of error = Critical value x Standard error of the statistic

Here critical value = 1.96

Standard error = (s/n) = 0.6/n

so 0.08 = 1.96 * 0.6/n

n = 1.96 * 0.6/ 0.08 = 14.7

n = 216.09

so n would be 216 or 217.

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