You have the following information:
average = x = 8.2
number of data sample = n = 90
standard deviation = σ = 2.5
In order to test the manufacturers claim, take into account that this situation is about hypothesis test for average value.
First, calculate the Z factor for the normal distribution, given by:
![Z=\frac{\bar{x}-\mu}{\sigma\text{ /}\sqrt[]{n}}](https://img.qammunity.org/2023/formulas/mathematics/college/kcki5ynxd26loq5ufek9c6orf9y1zhw1ts.png)
replace the values of the given parameters into the previous expression:
![\frac{9-8.2}{2.5/\sqrt[]{90}}=3.03](https://img.qammunity.org/2023/formulas/mathematics/college/w39gouqt8cvbdf10ct8nm3uqc22pv6e7pf.png)
Next, it is necessary to determine if such value is contained right side of the value of the normal distribution for 5%, which is the same that within an interval of 95% of confidence level.
By searching in a table of the normal distribution for a confidence level of %95. The value is:

As you can notice, the Z value obtained with the given parameters is greater than the Z value for a 95% confidence level.
Then, you can conclude that the manufacturer statment is false.