Final answer:
The 25th percentile is approximately 1191 chocolate chips, the middle 95% of bags have between 1138 and 1384 chocolate chips, and the interquartile range is 140 chocolate chips.
Step-by-step explanation:
We have a situation where the number of chocolate chips in a bag is approximately normally distributed with a mean (μ) of 1261 chips and a standard deviation (σ) of 117 chips. To address the questions, we will apply concepts from statistics related to the normal distribution.
25th Percentile
(a) To find the 25th percentile, also known as the first quartile (Q1), we would use a z-score table or a calculator with normal distribution function capabilities. The z-score for the 25th percentile in a standard normal distribution is approximately -0.675. We use the following formula to convert the z-score to the specific value:
X = μ + (z × σ)
So, we calculate:
X = 1261 + (-0.675 × 117) ≈ 1191 chocolate chips
Children's Percentiles
(b) The middle 95% of the bags' chocolate chip counts falls between the 2.5th and 97.5th percentiles. The corresponding z-scores are approximately ∑.96 and +1.96. We apply the same formula as before:
Lower bound X = μ + (-1.96 × σ) ≈ 1138 chocolate chips
Upper bound X = μ + (1.96 × σ) ≈ 1384 chocolate chips
Interquartile Range
(c) The interquartile range (IQR) is the difference between the third quartile (Q3) and the first quartile (Q1). The 75th percentile (Q3) can be found using the z-score for the 75th percentile, which is about 0.675. We then find Q3:
Q3 = μ + (0.675 × σ) ≈ 1331 chocolate chips
Therefore, the IQR is:
IQR = Q3 - Q1 = 1331 - 1191 = 140 chocolate chips