Question

Question 7 of 7 point Attempt 1 of 3 63 Section Exercise (cal) A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 50 families will be between 17 and 18 pounds. Assume that the sample is taken from a large population and the correction factor can be ignored. Use a TI-83 Plus/T1-84 Plus calculator and round the final answers to at least four decimal places. P(17<x<18)- dlo

Answer 1

7) Given, mean = 17.2 pounds, Standard deviation = 2.5 pounds

Probability that the mean of a sample of 50 will be between 17 and 18 pounds,

  $\small P(17\leq \bar{x}\leq 18)$

  $\small =P(\frac{17-17.2}{2.5/\sqrt{50}}\leq \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\leq \frac{18-17.2}{2.5/\sqrt{50}})$

  $\small =P(-0.566\leq Z\leq 2.263)$

= 0.7025

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