Answer:
the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444
Step-by-step explanation:
Given the data in the question;
x        f(x)         xp(x)               x²p(x)
1         1/3        0.33333        0.33333
2        1/3        0.66667        1.33333
3        1/3        1.00000        3.0000
∑                    2.0000          4.6667
∑(xp(x)) = 2
∑(x²p(x)) = 4.6667
Variance σ² = ∑(x²) - ∑(x)² = 4.6667 - (2)² = 4.6667 - 4 = 0.6667
standard deviation σ = √variance = √0.6667 = 0.8165
Now since, n = 33 which is greater than 30, we can use normal approximation
for normal distribution z score ( x-μ)/σ
mean μ = 2
standard deviation = 0.817
sample size n = 33 
standard of error σₓ = σ/√n = 0.817/√33 = 0.1422
so probability will be;
p( 2.1  < X < 2.4 ) = p(( 2.1-2)/0.1422) <  x"-μ/σₓ  <  p(( 2.4-2)/0.1422)
= 0.70 < Z < 2.81    
=  1 - ( 0.703 < Z < 2.812 )
FROM Z-SC0RE TABLE
=  1 - ( 0.25804 + 0.49752 )
= 1 - 0.75556
p( 2.1  < X < 2.4 ) = 0.2444
Therefore,  the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444