Respuesta :
Answer:
The percent of callers are 37.21 who are on hold.
Step-by-step explanation:
Given:
A normally distributed data.
Mean of the data, [tex]\mu[/tex] = 5.5 mins
Standard deviation, [tex]\sigma[/tex] = 0.4 mins
We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.
Lets find z-score on each raw score.
⇒ [tex]z_1=\frac{x_1-\mu}{\sigma}[/tex] ...raw score,[tex]x_1[/tex] = [tex]5.4[/tex]
⇒ Plugging the values.
⇒ [tex]z_1=\frac{5.4-5.5}{0.4}[/tex]
⇒ [tex]z_1=-0.25[/tex]
For raw score 5.5 the z score is.
⇒ [tex]z_2=\frac{5.8-5.5}{0.4}[/tex]
⇒ [tex]z_2=0.75[/tex]
Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.
We have to work with P(5.4<z<5.8).
⇒ [tex]P(5.4<z<5.8)[/tex]
⇒ [tex]P(-0.25<z<1.5)[/tex]
⇒ [tex]P(z<1.5)-P(z<-0.25)[/tex]
⇒ [tex]z(1.5)=0.7734[/tex] and [tex]z(-0.25)=0.4013[/tex]...from z -score table.
⇒ [tex]0.7734-0.4013[/tex]
⇒ [tex]0.3721[/tex]
To find the percentage we have to multiply with 100.
⇒ [tex]0.3721\times 100[/tex]
⇒ [tex]37.21[/tex] %
The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21
