We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.
The probability of a data score between two z-scores is given by formula
.
Using above formula, we will get:
![P(-1.32<z<-0.34)=P(z<-0.34)-P(z<-1.32)](https://img.qammunity.org/2021/formulas/mathematics/college/nghmaksye9qz6vfzeqiqlmrgzmyyl4c5cf.png)
Now we will use normal distribution table to find probability corresponding to both z-scores as:
![P(-1.32<z<-0.34)=0.36693-0.09342](https://img.qammunity.org/2021/formulas/mathematics/college/xvypy3kzu06tp5hrgor0d8fhf2u70sofvt.png)
![P(-1.32<z<-0.34)=0.27351](https://img.qammunity.org/2021/formulas/mathematics/college/6s7ft9az40ucvgw0dqhz8981v7zub8uusi.png)
Now we will convert
into percentage as:
![0.27351* 100\%=27.351\%](https://img.qammunity.org/2021/formulas/mathematics/college/2l1bddppytyxfdr9mum5s3g497trgrrdgl.png)
Upon rounding to nearest tenth of percent, we will get:
![27.351\%\approx 27.4\%](https://img.qammunity.org/2021/formulas/mathematics/college/mh72auakkwvltxwlo73svevxdd2jvstzc7.png)
Therefore, our required probability is 27.4% and option C is the correct choice.