Question

The income of a group of 10,000 persons was found to be normally distributed with means=rs.750 and standard deviation = rs.50. Show that of this group about 95% had income exceeding rs. 668 and only 5% had income exceeding rs. 832. What was the lowest income among the richest 100?

Answer 1

Answer:

Step-by-step explanation:

P((y - μ)/σ > - 1.6449) = 95%  

P(y - μ > - 1.6449σ) = 95%  

P(y > μ - 1.6449σ) = 95%  

P(y > 750 - 1.6449(50)) = 95%  

P(y > 667.755) = 95%  

P(y > 668) ≈ 95%  

P((y - μ)/σ > 1.6449) = 5%  

P(y - μ > + 1.6449σ) = 5%  

P(y > μ + 1.6449σ) = 5%  

P(y > 750 + 1.6449(50)) = 5%  

P(y > 832.245) = 5%  

P(y > 832) ≈ 5%

Answer 2

Answer:

AID13391: The income of a group of 10,000 persons was found to be normally distributed with mean Rs.750PM and standard deviation =Rs.50 show that of this group 95% had income exceeding Rs.668 and only 5% had income exceeding RS.832