Step 1 :The distribution of \(\bar{x}\) is approximately normal due to the Central Limit Theorem, as the sample size is large (n=36).
Step 2 :The mean of the distribution \(\mu_{\bar{x}}\) is equal to the mean of the population, which is 60.
Step 3 :The standard deviation of the distribution \(\sigma_{\bar{x}}\) is equal to the standard deviation of the population divided by the square root of the sample size, which is 21/\(\sqrt{36}\) = 3.5.
Step 4 :So, \(\bar{x}\) has a normal distribution with mean \(\mu_{\bar{x}}=60\) and standard deviation \(\sigma_{\bar{x}}=3.5\).
Step 5 :The z value is calculated by subtracting the mean from the x value and then dividing by the standard deviation. So, for \(\bar{x}=56.5\), the z value is (56.5-60)/3.5 = -1.
Step 6 :So, \(z=-1\).
Step 7 :The probability \(P(\bar{x}<56.5)\) is equivalent to the probability \(P(z<-1)\). Looking up -1 in the z-table, we find that the probability is approximately 0.1587.
Step 8 :So, \(P(\bar{x}<56.5)=0.1587\).
Step 9 :It would be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 56.5 because less than 5% of all such samples have means less than 56.5. The probability we calculated in part (c) is 0.1587, which is greater than 0.05, so more than 5% of all such samples have means less than 56.5.
Step 10 :Therefore, the correct answer is \(\boxed{\text{'No, it would not be unusual because more than 5% of all such samples have means less than 56.5.'}}\).