SOLUTION
The mean is 4min
standard deviation is 1min
the z score is
where
then we have
The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0
Hence, the percentage of the calls that lasted less than 3 min is 16%
Answer:
-32
Step-by-step explanation:
10–{22–[(−9)+(−11)]}
Work inside out
10–{22–[(-20)]}
Subtracting a negative is adding
10–{22+20}
10 - 42
-32
Your answer will be negative 2
The percentage of the scores that are between 75.8 and 89 is; 95%
<h3>How to find the percentage from z-score?</h3>
We are given;
Population mean; μ
Standard deviation; σ = 3.3
Thus;
z-score for a mean score of 75.8 is;
z = (75.8 - 82.4)/3.3
z = -2
z-score for a mean score of 89 is;
z = (89 - 82.4)/3.3
z = 2
From online p-value from z-score calculator, the p-value between both z-scores is;
p-value = 0.9545 = 95.45%
Approximating to the nearest percent = 95%
Read more about z-score at; brainly.com/question/25638875
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