A random variable follows a normal distribution with mean 44

A random variable follows a normal distribution with mean 44 and variance 16.

a. What is the probability of a randomly selected value falling between 40 and 48?

b. What is the 88% confidence interval for the mean if you had a sample of 55 values to determine the mean and standard deviation?

Solution

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    40      
x2 = upper bound =    48      
u = mean =    44      
          
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254 [ANSWER]
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492   [ANSWER]

b)

Note that              
              
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 =    0.06          
X = sample mean =    44          
z(alpha/2) = critical z for the confidence interval =    1.554773595          
s = sample standard deviation =    4          
n = sample size =    55          
              
Thus,              
              
Lower bound =    43.16141749          
Upper bound =    44.83858251          
              
Thus, the confidence interval is              
              
(   43.16141749   ,   44.83858251   ) [ANSWER]