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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%

2 Answers

4 votes

Answer:

C- 27.4%

Explanation:

User Tomision
by
5.6k points
6 votes

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* 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.

User Kakoni
by
5.8k points
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