Suppose x has a distribution with = 42 and = 9. (a) If random samples of size n

Question

Suppose x has a distribution with = 42 and = 9. (a) If random samples of size n = 16 are selected, can we say anything about the x distribution of sample means? Yes, the x distribution is normal with mean x = 42 and x = 0.6.Yes, the x distribution is normal with mean x = 42 and x = 2.25. Yes, the x distribution is normal with mean x = 42 and x = 9.No, the sample size is too small. (b) If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16? Yes, the x distribution is normal with mean x = 42 and x = 0.6.Yes, the x distribution is normal with mean x = 42 and x = 9. Yes, the x distribution is normal with mean x = 42 and x = 2.25.No, the sample size is too small. Find P(38 x 43). (Round your answer to four decimal places.)

Explanation / Answer

Suppose x has a distribution with = 42 and = 9. (a) If random samples of size n = 16 are selected, can we say anything about the x distribution of sample means?

A)

OPTION D: No, the sample size is too small.

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B)

If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16?

In this case, the sampling distribution of means will be normal regardless of sample size.

By central limit theorem, the sampling distribution of means have the same mean, u(X) = 42.

However, it has a reduced standard deviation,

sigma(X) = sigma/sqrt(n) = 9/sqrt(16) = 2.25.

Hence,

OPTION B: Yes, the x distribution is normal with mean x = 42 and x = 2.25. [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 =    38      
x2 = upper bound =    43      
u = mean =    42      
n = sample size =    16      
s = standard deviation =    9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.777777778      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.444444444      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.03772018      
P(z < z2) =    0.671639357      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.633919177   [ANSWER]  

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