The weights of packages filled by machine are normally distributed about a mean of 25 ounces, with a standard deviation of one ounce. What is the probability that n packages from the machine will have an average weight of less than 24 ounces if n = 1, 4, 16, 64? (10)
Mean, "u=25"
Standard deviation, "\\sigma=1"
Average weight, "\\bar{X}=24"
Case-1 When n=1
"P(z<24)=1-P(24<z<25) =1-\\dfrac{\\bar{X}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}=1-\\dfrac{24-25}{\\frac{1}{\\sqrt{1}}}=1-(-1)=2"
The value of z is 0.5080.
Case-2 When n=4
"P(z<24)=1-P(24<z<25) =1-\\dfrac{\\bar{X}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}=1-\\dfrac{24-25}{\\frac{1}{\\sqrt{4}}}=1-(-2)=3"
The value of z is 0.6852
Case -3 When n=16
"P(z<24)=1-P(24<z<25) =1-\\dfrac{\\bar{X}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}=1-\\dfrac{24-25}{\\frac{1}{\\sqrt{16}}}=1-(-4)=5"
The value of z is 0.8453
Case-4 When n=64,
"P(z<24)=1-P(24<z<25) =1-\\dfrac{\\bar{X}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}=1-\\dfrac{24-25}{\\frac{1}{\\sqrt{64}}}=1-(-8)=9"
The value of z is-0.9568
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Answer to Question #164892 in Statistics and Probability for Muhammad Fahad Fayyaz found incomplete and the method how get value of z is 0.5080 is not defined..
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