Answer:
The mean of this distribution is approximately 3.96. 
Step-by-step explanation:
Here's how to solve this problem using a normal distribution table.
Let  be the
 be the 
 .
.
In this question,  and
 and  . The equation becomes
. The equation becomes
 .
.
To solve for  , the mean of this distribution, the only thing that needs to be found is the value of
, the mean of this distribution, the only thing that needs to be found is the value of  . Since
. Since 
The problem stated that  . Hence,
. Hence,  .
.
The problem is that the normal distribution tables list only the value of  for
 for  . To estimate
. To estimate   from
 from  , it would be necessary to find the appropriate
, it would be necessary to find the appropriate 
Since  and is greater than
 and is greater than  ,
,  . As a result,
. As a result,  can be written as the sum of
 can be written as the sum of  and
 and  . Besides,
. Besides,  . As a result:
. As a result:
 .
.
Therefore:
 .
. 
Lookup  on a normal distribution table. The corresponding
 on a normal distribution table. The corresponding  -score is
-score is  . (In other words,
. (In other words,  .)
.) 
Given that
Solve the equation  for the mean,
 for the mean,  :
:
 .
.
 .
.