The mean is 442/9=49.1 approx.
The given mean of 57.5 by itself is insufficient, because we also need to know the spread of the data in order to judge if the sample mean is within the spread of the given mean. The spread is indicated by the standard deviation σ, so, given this value, we can judge whether 49.1 lies within 1.96 (95% degree of confidence) of these standard deviations from the mean of 57.5.
The difference between the means is 8.39 approx. We then use the root-mean-square of the standard deviations form the sample and population to calculate the standard deviation for the difference in means, but we don't have the population size.
However, we can do the complementary comparison by taking the standard deviation s of the sample=2.62 and seeing if 57.5 is significantly different.
The sample size is small so we need to use the t-distribution with 8 degrees of freedom to find a 95% critical value for the number of standard deviations from the mean. This critical value is 2.306. And s has to be adjusted to s/√9=2.62/3=0.87 approx. So the deviation from the mean is 0.87×2.306=2 approx.
A confidence interval then is 49.1±2 or [47.1,51.1]. 57.5 is not within this interval, so the sample mean is significantly different from 57.5.