Answer:
6.55%
Explanation:
Given a random variable X having a normal distribution with:
• Mean, µ = 40
,
• Standard Deviation, σ = 10
We want to find the probability that X is between 55 and 70.
In order to do this, first, we find the z-scores for X=55 and X=70.
The z-score formula is:
![z=(X-\mu)/(\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/2eurhv0e2l78yy8nqvl2450glsjqpn08m2.png)
Thus:
![\begin{gathered} At\text{ X=55, }z=(55-40)/(10)=(15)/(10)=1.5 \\ At\text{ X=70, }z=(70-40)/(10)=(30)/(10)=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gclvu5qt60g8x84kfgnn8kicy2sz8ao1mn.png)
From the z-table:
![\begin{gathered} P\mleft(x<1.5\mright)=0.93319 \\ P\mleft(x>3\mright)=0.0013499 \\ P\mleft(x<1.5orx>3\mright)=0.93319+0.0013499=0.93454 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/98xxtbg6dj2dkz2zbs7fpqoi9x736xxxk2.png)
Therefore, using a z-score table, the probability that X assumes a value between 55, 70 is:
![\begin{gathered} P\mleft(1.53) \\ =1-0.93454 \\ =0.065457 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/luw0q57g4mz40gniworr53r9f5rg8btj2f.png)
The required probability is 6.55%.