Statistics?

Question

According to a survey of the top 10 employers in a major city in the Midwest, a worker spends an average of 413 minutes a day on the job. Suppose the standard deviation is 26.8 minutes and the time spent is approximately a normal distribution. What are the times that approximately 95.45% of all workers will fall?

Answer 1

Mean of normal distribution is (µ) 413 min

Standard deviation is (σ) 26.8

Number of employers (n) = 10

According to three-sigma rule or empirical rule μ ± 2σ  = 0.9545

Standard deviation of x distribution

= 8.474

95.45% of all workers will fall in μ ± 2

= 413 - 2×8.474 to 413 + 2×8.474

= 396.056 min to 429.94 min

So 95.45% of all workers will fall in 396.056 min to 429.94 min