A random sample of 15 employees was selected. The average age in the sample was 31 years with a variance of 49 years. Assuming ages are normally distributed, the 98% confidence interval for the population average age is . a. 26.26 to 35.74 b. 11.54 to 18.46 c. 25.62 to 36.38 d. 27.82 to 34.18

Option a : 26.26 to 35.74 .

Step-by-step explanation:

We are provided a random sample of 15 employees of which average age in the sample, xbar = 31 years and Standard Deviation, s = = 7 years.

We know that   follows

So, 98% confidence interval is given by ;

   P(-2.624 < < 2.624) = 0.98 {because at 14 degree of freedom t table

                                                        gives critical value of 2.624 at 1% level}

  P(-2.624 < < 2.624) = 0.98

  P(-2.624* < < 2.624* ) = 0.98

  P(xbar - 2.624* < < xbar + 2.624* ) = 0.98

98% Confidence Interval for = [xbar - 2.624* , xbar + 2.624* ]

                                                  = [ , ]

                                                  = [26.26 , 35.74]

Therefore, 98% confidence interval for the population average age is 26.26 to 35.74 .

the probability is 0

i dont know sorry

step-by-step explanation: