81.8% percent of the numbers fall between 35 and 50.
According to the question
Mean = [tex]\mu[/tex] = 40
Standard deviation = [tex]\sigma[/tex] = 5
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the z-score of a measure X is given by:
z = [tex]\frac{x-\mu}{\sigma}[/tex]
Now we are supposed to find out what percent of the numbers fall between 35 and 50
z = [tex]\frac{x-\mu}{\sigma}[/tex]
Substitute the values
z = [tex]\frac{x-40}{5}[/tex]
Now for P(35<x<50)
Substitute x = 35
z = [tex]\frac{35-40}{5}[/tex]
z = -1
Substitute x = 50
z = [tex]\frac{50-40}{5}[/tex]
z = 2
So, P(-1<z<2)
P(z<2) - P(z<-1)
= 0.9772 - 0.1587
= 0.8185
= 0.818 × 100
= 81.8%
Hence, 81.8% percent of the numbers fall between 35 and 50.
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