[tex]X[/tex] is the random variable for lifespan of a protozoan and [tex]X\sim\mathcal N(48,10.2^2)[/tex]. Let [tex]\bar X[/tex] be the mean of a sample from this distribution, so that [tex]\bar X\sim\mathcal N\left(48,\left(\dfrac{10.2}{\sqrt{49}}\right)^2\right)[/tex].
For the sake of clarity, I'm denoting a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] by [tex]\mathcal N(\mu,\sigma^2)[/tex].
We have
[tex]P(\bar X\ge49)=P\left(\dfrac{\bar X-48}{\frac{10.2}{\sqrt{49}}\ge\dfrac{49-48}{\frac{10.2}{\sqrt{49}}\right)\approx P(Z\ge0.6863)\approx0.2463[/tex]
(where [tex]Z[/tex] follows the standard normal distribution)
