# statistics question 10 point reward !?

small local company were asked what their annual salary was. For this group was \$22,500, with a standard deviation of \$1500.

96 confidence interval for the average annual salary of all employee of this company?

Relevance

X : salary of employees

xmean = 22500

sx = 1500

Pf(22500 - 1500*2.055< μx < 22500 + 1500*2.055) = 0.96

Pf( 19417.5 < μx < 25582.5 ) = 0.96

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• Look up a table of the single tail normal distribution to see how many standard deviations away from the mean will include 48% of the values. Then multiply this by 2 x \$1500.

Hint: 2 standard deviations from the mean includes 47.7 % of the data on one side, or 95.4% of all the data for both tails, so your answer should be just a little over \$3000

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• Look up your z-table for 0.96 to get the z value (should be about 2). The 96% CI will be:

22,500 +/- (1500 * z-value).

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• approx 2.0545 x 1500...so 22500 +- 3081.75 is the 96% range of salaries.

Source(s): Me
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