The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The width of a 98 percent CI for the true mean client ageis approximately:
A. ± 1.711 years.
B. ± 2.326 years.
C.± 2.492 years.
D. ± 2.797 years.
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Ответ:
its 200
Explanation:
600-100=500-2=498 600+100=700-2=798 798-498= 200