Answer: Option 'A' is correct.
Step-by-step explanation:
Since we have given that
Population Mean weight (\(\mu\))= 10 grams a piece
Standard deviation of the weight of a box = 3 grams
Number of mints = 10
We need to buy a box of mints that weighs 95 grams.
Sample mean is given by
\(x=\frac{95}{10}=9.5\ grams\)
First we find out the standard error which is given by
\(s=\frac{\sigma}{\sqrt{n}}\\\\=\frac{3}{\sqrt{10}}\\\\=0.94868\)
Since it is normal distribution, so, we will find z-score.
\(z=\frac{x-\mu}{s}\\\\z=\frac{9.5-10}{0.94868}\\\\z=-0.527\\\\z=-0.53\)
The area to the left of a z-score of -0.53 = 0.29805.
So, it may be 90% or 95 % confidence.
For 95% confidence level,
\(\alpha=\frac{1-0.95}{2}=0.025\)
Similarly,
For 90% confidence level,
\(\alpha=\frac{1-0.90}{2}=0.05\)
We have little confidence that the box he bought did not come from the factory. that is much smaller than 0.05.
So, it is safe to assume 90% confidence.
So, we will get 90% confidence, critical value = 1.645
Margin of error is given by
\((Standard\ deviation)\times (critical\ value)\\\\=0.94868\times 1.645\\\\=1.56\)
So, confidence interval will be
(10-1.56,10+1.56)
=(8.44,11.56)
Hence, Option 'A' is correct.
