The document discusses how to calculate the necessary sample size 'n' for a uniform distribution with mean 4 and standard deviation 2.31 to ensure that the probability P(3.5 ≤ x̅ ≤ 4.5) is at least 0.8. It concludes that the required sample size 'n' must be approximately 35.

1. Example
A uniform distribution has density function
𝑓(𝑥) =
1
8
0 ≤ 𝑥 ≤ 8
Find how large the value of ‘n’ must be in order
𝑃(3.5 ≤ 𝑥̅ ≤ 4.5) ≥ 0.8
Solution
First to find mean and standard deviation of uniform
distribution.
𝑀𝑒𝑎𝑛 𝑜𝑓 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 =
𝑎 + 𝑏
2
=
0 + 8
2
= 4
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 =
√
(𝑏−𝑎)2
12
= √
(8−0)2
12
= 2.31
As the Z- distribution is symmetrical about mean
therefore the area before and after the mean will be equal.
That is 𝑃(3.5 ≤ 𝑥̅ ≤ 4.5) ≥ 0.8

2. 𝑃 (
3.5−𝜇
𝜎
√𝑛
⁄
≤
𝑥̅−𝜇
𝜎
√𝑛
⁄
≤
4.5−𝜇
𝜎
√𝑛
⁄
) ≥ 0.8
𝑃 (
3.5−4
2.31
√𝑛
⁄
≤
𝑥̅−𝜇
𝜎
√𝑛
⁄
≤
4.5−4
2.31
√𝑛
⁄
) ≥ 0.8
The value of Z at 𝑃(𝑥̅ ≤ 3.5) = 0.1000 𝑖𝑠 − 1.28
By comparing
3.5 − 4
2.31
√𝑛
⁄
= −1.28
𝑛 = (
−1.28 × 2.31
−0.5
)
2
𝑛 = 34.97 ≃ 35