Answer:
[tex]X ≤ P(119) \rightarrow P(X < 119.5)[/tex]
Step-by-step explanation:
Given:
n = 504
p = 0.27
To Approximate:
P(x≤119) usimg normal distribution
Solution:
We know that,
[tex]p+q = 1\\q = 1-p\\q = 1 - 0.27 \\ \fbox{q = 0.73}[/tex]
let's find the mean,
[tex]n \cdot p = 500 \times 0.27 \\ n \cdot p = 135 \\ \mu \: = 135[/tex]
let's calculate standard deviation,
[tex] \sigma = \: \sqrt {n \cdot p\cdot q} \\ \sigma \: = \sqrt{504 \times 0.27 \times 0.73} \\ \sigma \: = 9.96[/tex]
Rewriting the equation using continuity correction,
[tex] P(X ≤ n) \rightarrow P(X < n + 0.5) \\ P(X ≤ 119) \rightarrow P(X < 119 + 0.5) \\ P(X ≤ 119) \rightarrow P(X < 119.5) \\ \fbox{x \: = 119.5}[/tex]
Some more things you must need to know,
standard score is an important part of statistics,
can be derived using formula
[tex]Z = \frac{ x - \mu}{ \sigma} \\ Z = \frac{ 119.5 - 135}{ 9.96} \\ Z= \frac{-15.5}{9.96} \\ \fbox{Z= -1.55}[/tex]
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