Question

Please explain steps to find these answers on the TI 84 plus calculator:

A study found that the mean migration distance of the green turtle was 2200 kilometers and the standard deviation was 625 kilometers. Assuming that the distances are normally distributed, find the probability that a randomly selected green turtle migrates a distance of

            a. Less than 1900 kilometers

            b. Between 2000 kilometers and 2500 kilometers

            c. Greater than 2450 kilometers

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	2200
std deviation   =σ=	625

probability =

P(X<1900)

=

P(Z<-0.48)=

0.3156

(for Ti-84 : go to 2nd vars , 2:normal cdf : lower : -20000, upper: 1900 , mu=2200 , sigma =625 ,paste , )

b)

probability =

P(2000<X<2500)

=

P(-0.32<Z<0.48)=

0.6844-0.3745=

0.3099

(for Ti-84 : go to 2nd vars , 2:normal cdf : lower : 2000, upper: 2500 , mu=2200 , sigma =625 ,paste , )

c)

probability =

P(X>2450)

=

P(Z>0.4)=

1-P(Z<0.4)=

1-0.6554=

0.3446

(for Ti-84 : go to 2nd vars , 2:normal cdf : lower : 2450, upper: 10000 , mu=2200 , sigma =625 ,paste , )