Answer:
Standard deviation σ = 1.40 (2 d.p.)
Explanation:
If a continuous random variable X is normally distributed with mean μ and variance σ², it is written as:
![\boxed{X \sim\text{N}(\mu,\sigma^2)}](https://img.qammunity.org/2023/formulas/mathematics/college/tcdnop5lgvnwub5rexij3dnat1thy01fhh.png)
Given:
- Mean μ = 13.9 g/dL
- P(X < 16.2) = 0.95
Therefore, if the haemoglobin levels are normally distributed:
![X \sim\text{N}(13.9,\sigma^2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/wky8h6latcp20hutabumpz1fikzlo98fg8.png)
where X is the haemoglobin level.
Converting to the Z distribution:
![\boxed{\textsf{If }\: X \sim\textsf{N}(\mu,\sigma^2)\:\textsf{ then }\: (X-\mu)/(\sigma)=Z, \quad \textsf{where }\: Z \sim \textsf{N}(0,1)}](https://img.qammunity.org/2023/formulas/mathematics/college/lbnf63vdeyoeq7uhs9v0g99wqna5dkh2z1.png)
Transform X to Z:
![\text{P}(X < 16.2)= \text{P}\left(Z < (16.2-13.9)/(\sigma)\right)= 0.95](https://img.qammunity.org/2023/formulas/mathematics/high-school/ae5ux7s783tgp82j17pqkeosncngarxrgg.png)
According to the z-tables, when p = 0.95, z = 1.6449
![\implies (16.2-13.9)/(\sigma)=1.6449](https://img.qammunity.org/2023/formulas/mathematics/high-school/5u7oqyafmuzjy34poddg8lmcbnoyncmhok.png)
![\implies (2.3)/(\sigma)=1.6449](https://img.qammunity.org/2023/formulas/mathematics/high-school/r5ha7gq0l7b4kpwjnoa58r0k3h694aqeqw.png)
![\implies \sigma=(2.3)/(1.6449)](https://img.qammunity.org/2023/formulas/mathematics/high-school/upfdzs2te0li9ypy23l2tdy7nyhdhs4snj.png)
![\implies \sigma=1.40](https://img.qammunity.org/2023/formulas/mathematics/high-school/dhr3b1wiq7p0uiguuv1pfsvbque92ykua9.png)