Answer:
[tex]P(0.7<Z<1.4)[/tex]
And we can find this probability with the following difference:
[tex]P(0.7<Z<1.4)= P(Z<1.4)- P(Z<0.7)[/tex]
We can use the following commands on the ti 84
2nd>VARS>DISTR
And then we look for normalcdf and we input this:
normalcdf(0.7,1.4,0,1)
The other possible code would be:
normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)
And we got:
[tex]P(0.7<Z<1.4)=0.161[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
For this case we want to find this probability:
[tex]P(0.7<Z<1.4)[/tex]
And we can find this probability with the following difference:
[tex]P(0.7<Z<1.4)= P(Z<1.4)- P(Z<0.7)[/tex]
We can use the following commands on the ti 84
2nd>VARS>DISTR
And then we look for normalcdf and we input this:
normalcdf(0.7,1.4,0,1)
The other possible code would be:
normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)
And we got:
[tex]P(0.7<Z<1.4)=0.161[/tex]
Answer: 0.1612
Step-by-step explanation:
P(0.7 <= Z <= 1.4)= P(Z<=1.4)- P(<=0.7)
= 0.91924 - 0.75804
= 0.1612
