Question

What is 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 Round your answer to the nearest tenth of a percent!

A- 26.4%

B- 28.4%

C- 27.4%

D- 29.4%

Answer 1

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 $P(a<z<b)=P(z<b)-P(z<a)$.

Using above formula, we will get:

$P(-1.32<z<-0.34)=P(z<-0.34)-P(z<-1.32)$

Now we will use normal distribution table to find probability corresponding to both z-scores as:

$P(-1.32<z<-0.34)=0.36693-0.09342$

$P(-1.32<z<-0.34)=0.27351$

Now we will convert $0.27351$ into percentage as:

$0.27351\times 100\%=27.351\%$

Upon rounding to nearest tenth of percent, we will get:

$27.351\%\approx 27.4\%$

Therefore, our required probability is 27.4% and option C is the correct choice.

Answer 2

Answer:

C- 27.4%

Step-by-step explanation: