Given a normal population whose mean is 50 and whose standard deviation is 5, find the probability that a random sample of a. 4 has a mean between 49 and 52. b. 16 has a mean between 49 and 52. c. 25 has a mean between 49 and 52
"\\mu=50 \\\\\n\n\\sigma=5"
a. n=4
"P(49<\\bar{x}<52) = P(\\bar{x}<52) -P(\\bar{x}<49) \\\\\n\n=P(Z< \\frac{52-50}{5 \/ \\sqrt{4}}) -P(Z< \\frac{49-50}{5 \/ \\sqrt{4}}) \\\\\n\n=P(Z< 0.8) -P(Z< -0.4) \\\\\n\n= 0.7881 - 0.3445 \\\\\n\n=0.4436"
b. n=16
"P(49<\\bar{x}<52) = P(\\bar{x}<52) -P(\\bar{x}<49) \\\\\n\n=P(Z< \\frac{52-50}{5 \/ \\sqrt{16}}) -P(Z< \\frac{49-50}{5 \/ \\sqrt{16}}) \\\\\n\n=P(Z< 1.6) -P(Z< -0.8) \\\\\n\n= 0.9452 - 0.2118 \\\\\n\n=0.7334"
c. n=25
"P(49<\\bar{x}<52) = P(\\bar{x}<52) -P(\\bar{x}<49) \\\\\n\n=P(Z< \\frac{52-50}{5 \/ \\sqrt{25}}) -P(Z< \\frac{49-50}{5 \/ \\sqrt{25}}) \\\\\n\n=P(Z< 2) -P(Z< -1) \\\\\n\n= 0.9772 - 0.1586 \\\\\n\n=0.8186"
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