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
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