AP Statistics 2.11 The Normal Distribution- Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
The top 10% corresponds to the 90th percentile of the normal distribution.
Using the standard normal table:
\(z_{0.90} \approx 1.28\)
Convert this z-score to the original scale:
\(x=\mu+z\sigma\)
\(x=86+(1.28)(2.7)\)
\(x=86+3.456\)
\(x\approx 89.46\)
Therefore, approximately 10% of the pitches are faster than 89.46 mph, making choice (D) correct.
✅ Answer: (D)
▶️ Answer/Explanation
A cookie is recycled if its weight is more than 0.4 grams away from the mean of 3.1 grams. Therefore, we want:
\(P(|X-3.1|>0.4)\)
Convert 0.4 grams to a z-score:
\(z=\frac{0.4}{0.25}=1.6\)
Thus we need the probability outside the interval:
\(-1.6<Z<1.6\)
Using the standard normal distribution:
\(P(Z<1.6)\approx0.9452\)
\(P(Z<-1.6)\approx0.0548\)
So the probability within 1.6 standard deviations is:
\(0.9452-0.0548=0.8904\)
The probability of being recycled is the probability in both tails:
\(1-0.8904=0.1096\)
\(\approx0.11\)
Therefore, the probability that a randomly selected cookie will be recycled is approximately 0.11.
✅ Answer: (B)
▶️ Answer/Explanation
The manufacturer wants only 1% of bags to be below 12 oz. Therefore, 12 oz. should correspond to the 1st percentile of the Normal distribution.
The z-score for the 1st percentile is approximately:
\( z = -2.33 \)
Using the z-score formula:
\( z = \frac{x-\mu}{\sigma} \)
\( -2.33 = \frac{12-\mu}{0.08} \)
Solving for \( \mu \):
\( 12-\mu = -2.33(0.08) \)
\( 12-\mu = -0.1864 \)
\( \mu = 12.1864 \)
Rounding to the nearest hundredth:
\( \mu \approx 12.19 \)
Therefore, the machine should be set to a mean filling weight of 12.19 oz. so that at most 1% of bags are underweight.
✅ Answer: (D)