Question

Suppose a population is known to be normally distributed with a mean, μ, equal to 116 and a standard deviation, σ, equal to 14. Approximately what percent of the population would be between 102 and 130?

Answer 1

Solution :

Given that ,

mean = = 116

standard deviation = = 14   

P(102< x <130 ) = P[(102-116) /14 < (x - ) / < (130-116) / 14)]

= P( -1< Z < 1)

= P(Z <1 ) - P(Z < -1)

Using z table   

= 0.8413 -0.1587

=0.6826

=68.26%