A random sample is drawn from a population with mean = 74 and standard deviation

Question

A random sample is drawn from a population with mean = 74 and standard deviation = 6.2. Use Table 1.

Is the sampling distribution of the sample mean with n = 18 and n = 47 normally distributed?

Can you use the standard normal distribution to calculate the probability that the sample mean falls between 74 and 77 for both sample sizes?

Calculate the above probability for n = 47. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

A random sample is drawn from a population with mean = 74 and standard deviation = 6.2. Use Table 1.

Explanation / Answer

Usually, we treat the sampling distribution when n > 30 normally distributed. If you use a different standard, please use that.

A)

OPTION D: No, only the sample mean with n = 47 will have a normal distribution. [ANSWER, D]

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b)

OPTION D: No, only for the sample size with n = 47, standard normal distribution could be used. [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    74      
x2 = upper bound =    77      
u = mean =    74      
n = sample size =    47      
s = standard deviation =    6.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    0      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3.32      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.5      
P(z < z2) =    0.999549913      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.499549913   [ANSWER]  

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