1. A random sample of 16 observations is to be drawn from a normal distribution having mean 11 and standard deviation 3. Let denote the sample mean. Find, correct to three decimal places
(i) The probability that will have a value between 9.2 and 12.2
(ii) The value of c for which
Solution:
(a) Given that,
mean "=\\mu=11"
standard deviation "=\\sigma=3"
"n=16 \n\\\\\\mu_{\\bar{x}}=11 \n\\\\\\sigma_{\\bar{x}}=\\sigma \/ \\sqrt n=3 \/ 4=0.75\n\\\\ \\left.\\mathrm{P}(9.2<\\bar{x}<12.2)=\\mathrm{P}\\left[(9.2-11) \/ 0.75<\\left(\\bar{x}-\\mu_{\\bar{x}}\\right) \/ \\sigma_{\\bar{x}}<(12.2-11) \/ 0.75\\right)\\right] \n\\\\=\\mathrm{P}(-2.4<Z<1.6) \n\\\\=\\mathrm{P}(Z<1.6)-\\mathrm{P}(Z<-2.4)"
Using z table
=0.9452-0.0082
probabilty =0.9370
(b) To find The value of c for which "P(\\bar{x}<c)=0.03".
"P(\\bar{x}<c)=0.03"
Using standard normal table,
"\\begin{aligned}\n\n&P(Z<z)=0.03 \\\\\n\n&=P(Z<-1.88)=0.03\n\n\\end{aligned}"
z = -1.88 Using standard normal z table,
Using z-score formula
"\\begin{aligned}\n\n&\\bar{x}=\\mathrm{z} \\times \\sigma_{\\bar{x}}{+} \\mu_{\\bar{x}} \\\\\n\n&\\bar{x}=-1.88 \\times 0.75+11 \\\\\n\n&\\bar{x}=9.59\n\n\\end{aligned}"
So, the value of c =9.59
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