= Solution
Allow each subject to have a separate baseline and slope through a <correlated random-intercept and random-slope model>:
$$
Y_{ij}=\beta_0+\beta_1t_j+b_{0i}+b_{1i}t_j+\varepsilon_{ij},\qquad
\begin{pmatrix}b_{0i}\\b_{1i}\end{pmatrix}\overset{\mathrm{iid}}\sim
N_2\left(0,
\begin{pmatrix}\tau_0^2&\rho\tau_0\tau_1\\\rho\tau_0\tau_1&\tau_1^2\end{pmatrix}\right),
\qquad \varepsilon_{ij}\overset{\mathrm{iid}}\sim N(0,\sigma^2).
$$
The subject <random effects> are independent of all measurement errors, and subjects are independent. The within-person <covariance> between times $s,t$ is $\tau_0^2+(s+t)\rho\tau_0\tau_1+st\tau_1^2$, with an additional $\sigma^2$ when the same reading is used twice.
\b[Prefer the model with a <random slope>.] The <maximum likelihood estimation> refits improve twice the <log-likelihood> by $227.9604$ while adding two <covariance> <statistical parameters>. Both the <Akaike information criterion> ($5165.779$ to $4941.819$) and <Bayesian information criterion> ($5183.367$ to $4968.200$) strongly favour it. The printed nominal <likelihood-ratio test> is overwhelming. The usual reference <chi-squared distribution> with two <statistical degrees of freedom> is not an exact regular calibration, since zero slope <variance> is a boundary and the intercept–slope <correlation> is then unidentified; a design-specific <parametric bootstrap> could calibrate it. This qualification does not undermine the substantial descriptive improvement shown by both information criteria.
The preferred fit estimates population mean baseline strength $\widehat\beta_0=61.09286$ kg and weekly gain $\widehat\beta_1=2.83143$ kg/week. The between-person baseline <standard deviation> is $\widehat\tau_0=15.785582$ kg, and the between-person slope <standard deviation> is $\widehat\tau_1=2.738666$ kg/week. Their estimated <correlation> $\widehat\rho=0.117$ is weakly positive, giving random-effect <covariance> about $5.058$ kg$^2$/week; its uncertainty is not supplied. The measurement-error <standard deviation> is $\widehat\sigma=9.923280$ kg. The printed $-0.038$ instead describes the <correlation> between the estimated <fixed effects>, not between the subject <random effects>. These <statistical parameter> estimates come from <restricted maximum likelihood>; the model comparison uses the separate <maximum likelihood estimation> fits.
The population mean fitted trajectory is $61.09286+2.83143t$ kg. Hence
$$
\boxed{\text{mean ten-week gain}=10\widehat\beta_1=28.3143\ \mathrm{kg},
\qquad \text{mean at week ten}=89.40716\ \mathrm{kg}.}
$$
Using the reported slope <standard error>, an approximate 95% <confidence interval> for the population mean gain is $28.3143\pm1.96(2.984464)=(22.46,34.16)$ kg; a <Student t confidence interval> with 499 <statistical degrees of freedom> is almost identical.
Successful individuals can improve far more than the average. Their latent ten-week gains have fitted <normal distribution>
$$
G_i=10(\beta_1+b_{1i})\sim N(28.3143,27.38666^2).
$$
A clear estimate for an upper-performing group uses a <between-person slope quantile>: the 95th percentile is $28.3143+1.645(27.38666)\simeq73.4$ kg, and the 97.5th percentile is about $82.0$ kg. \b[The upper 5% of fitted underlying gains begin around 73 kg.] These are person-to-person performance <quantiles>, not <confidence intervals> for the population mean. Identifying the best observed trainee would require that person's data or fitted subject <random effects>; the aggregate output cannot identify a literal maximum. If performance means the observed difference of endpoint readings, add the measurement-error <variance> $2\sigma^2$ to the latent gain <variance>.
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