Answer:
The probability that the mean of my sample will be between 24 and 25 cm
P(24 ≤X⁻≤25) = 0.4772
Step-by-step explanation:
Step(i):-
Given mean of the Population 'μ'= 25c.m
Given standard deviation of the Population 'σ' = 8c.m
Given sample size 'n' = 256
Let X₁ = 24
[tex]Z_{1} = \frac{x_{1}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{24-25}{\frac{8}{\sqrt{256} } } = -2[/tex]
Let X₂ = 25
[tex]Z_{2} = \frac{x_{2}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{25-25}{\frac{8}{\sqrt{256} } } = 0[/tex]
Step(ii):-
The probability that the mean of my sample will be between 24 and 25 cm
P(24 ≤X⁻≤25) = P(-2≤ Z ≤0)
= P( Z≤0) - P(Z≤-2)
= 0.5 + A(0) - (0.5- A(-2))
= A(0) + A(2) ( ∵A(-2) =A(2)
= 0.000+ 0.4772
= 0.4772
Final answer:-
The probability that the mean of my sample will be between 24 and 25 cm
P(24 ≤X⁻≤25) = 0.4772