Let "X=" the time needed to complete a final examination: "X\\sim N(\\mu, \\sigma^2)."
Then "Z=\\dfrac{X-\\mu}{\\sigma}\\sim N(0, 1)"
Given "\\mu=61, \\sigma=9"
1.1
"\\approx0.889188"
1.2
"=P(Z\\leq\\dfrac{80-61}{9})-P(Z<\\dfrac{75-61}{9})="
"=P(Z\\leq\\dfrac{19}{9})-P(Z<\\dfrac{14}{9})\\approx"
"\\approx0.9826186-0.9400931\\approx0.042526"
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A research firm conducted a survey to determine the mean amount of money smokers spend on cigarettes during a day. A sample of 100 smokers revealed that the sample mean is N$ 5.24 and sample standard deviation is N$ 2.18 Assume that the sample was drawn from a normal population. 2.1 Find the point estimate of the population mean. 2.2 Determine the lower limit of the 95% conference interval for estimating the unknown population mean
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