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:
[tex]Z=\frac{\bar{x}-\mu}{\sigma\text{ /}\sqrt[]{n}}[/tex]replace the values of the given parameters into the previous expression:
[tex]\frac{9-8.2}{2.5/\sqrt[]{90}}=3.03[/tex]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:
[tex]Z=1.65[/tex]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.