In a sample or 10 randomly selected women it was found that

In a sample or 10 randomly selected women, it was found that their mean heignt was 63. 4 inches. From previous studies. It is assumed fiat the standard deviation sigma 2.4 and that the population of height measurements is normally distributed. Construct the 95% confidence interval for the population mean (58.1, 67.3) (59.7, 66.5) (60.8, 65.4) 61.9, 64.9)

Solution

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    63.4          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2.4          
n = sample size =    10          
              
Thus,              
Margin of Error E =    1.487508078          
Lower bound =    61.91249192          
Upper bound =    64.88750808          
              
Thus, the confidence interval is              
              
(   61.9   ,   64.9   ) [answer, D]