Please hellppppp...................
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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, = 5.5 mins
Standard deviation, = 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.
⇒ ...raw score,
=
⇒ Plugging the values.
⇒
⇒
For raw score 5.5 the z score is.
⇒
⇒
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).
⇒
⇒
⇒
⇒ and
.<em>..from z -score table.</em>
⇒
⇒
To find the percentage we have to multiply with 100.
⇒
⇒ %
The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21