‘in the fall harvest are normally distributed, with mean weight of 200 The weights of apples grams and standard deviation of 8 grams What is the percentage of apples between 184 and 216 grams? 12 (a) 50% (0) 682 (c) 47.5% (d) 34% (e) 95%’
The Correct Answer and Explanation is:
To solve this, we can use the properties of the normal distribution. Given that the weights of apples follow a normal distribution with:
- Mean (μ) = 200 grams
- Standard deviation (σ) = 8 grams
We need to find the percentage of apples that weigh between 184 grams and 216 grams.
Step 1: Standardize the values using the Z-score formula
The Z-score formula is: Z=X−μσZ = \frac{X – \mu}{\sigma}Z=σX−μ
Where:
- XXX is the data point,
- μ\muμ is the mean,
- σ\sigmaσ is the standard deviation.
Now, calculate the Z-scores for 184 grams and 216 grams.
For 184 grams:
Z=184−2008=−168=−2Z = \frac{184 – 200}{8} = \frac{-16}{8} = -2Z=8184−200=8−16=−2
For 216 grams:
Z=216−2008=168=2Z = \frac{216 – 200}{8} = \frac{16}{8} = 2Z=8216−200=816=2
Step 2: Use the Z-table (standard normal distribution table) to find the probabilities
For Z = -2 and Z = 2, we look up the corresponding probabilities from the Z-table:
- The probability for Z = -2 is approximately 0.0228.
- The probability for Z = 2 is approximately 0.9772.
Step 3: Calculate the percentage between 184 and 216 grams
To find the percentage of apples that fall between 184 grams and 216 grams, we subtract the probability at Z = -2 from the probability at Z = 2: P(184≤X≤216)=P(Z=2)−P(Z=−2)=0.9772−0.0228=0.9544P(184 \leq X \leq 216) = P(Z = 2) – P(Z = -2) = 0.9772 – 0.0228 = 0.9544P(184≤X≤216)=P(Z=2)−P(Z=−2)=0.9772−0.0228=0.9544
Step 4: Convert to percentage
To convert this to a percentage: 0.9544×100=95.44%0.9544 \times 100 = 95.44\%0.9544×100=95.44%
Conclusion:
The percentage of apples weighing between 184 grams and 216 grams is approximately 95%. Therefore, the correct answer is:
(e) 95%
