Question

According to a recent​ study, the carapace length for adult males of a certain species of tarantula are normally distributed with a mean of μ=17.24mm and a standard deviation of σ=1.68 mm. Find the percentage of the tarantulas that have a carapace length between 15 mm and 16 mm.

Answer 1

Solution :

Given that ,

mean = = 17.24

standard deviation = = 1.68

P(15< x < 16) =P[(15-17.24) /1.68 < (x - ) / < (16-17.24) /1.68 )]

= P(-1.33 < Z < -.0.74)

= P(Z < -0.74) - P(Z <-1.33 )

Using z table,  

= 0.2296-0.0918

=0.1378

=13.78%

percentage of the tarantulas that have a carapace length between 15 mm and 16 mm. 13.78%