This question evaluates z-score application to heights in normal distributions for AP Statistics. Heights are normal with $\mu = 69$ inches and $\sigma = 3$ inches. For 63 inches, $z = \frac{63 - 69}{3} = -2$, meaning 2 standard deviations below the mean. The empirical rule notes 95% within $\pm 2\sigma$, so this is unusually short. Choice E is a distractor, wrongly claiming most are shorter, but only about 2.5% are in that lower tail. Normal models help by standardizing values to compare across distributions, emphasizing how $\sigma$ quantifies deviation from $\mu$.

This question assesses interpreting lifetimes in normal distributions via AP Statistics. Bulb lifetimes are normal with μ = 1200 hours and σ = 150 hours. The z-score for 1050 hours is (1050 - 1200) / 150 = -1, or 1 standard deviation below the mean. The empirical rule shows 68% within ±1σ, so this is somewhat short but typical, not unusual. Distractor choice E incorrectly states 95% last less than 1050, confusing the rule—actually, about 16% are shorter in the lower tail. In normal models, we reference μ as the center and σ for spread to understand proportions and typical ranges.

This question tests positioning a song length in a normal distribution with μ = 3.8 minutes and σ = 0.6. For 2.6, z = (2.6 - 3.8) / 0.6 = -2, 2 standard deviations below the mean, unusual in the left tail with about 2.5% shorter songs. This marks it as notably short. Choice C distracts by saying 2 SDs above, ignoring the subtraction direction. Mini-lesson: Normal curves are bell-shaped with most data near μ; use z = (x - μ) / σ to classify: central for |z| < 1, tails for |z| > 2, for intuitive rarity judgments.

This question tests z-score calculation for blood pressure readings. Given μ = 120 mmHg and σ = 15 mmHg, we find the z-score for x = 135 mmHg: z = (135 - 120)/15 = 15/15 = 1. A blood pressure of 135 mmHg is exactly 1 standard deviation above the mean, making it higher than average but not especially unusual. The distractors include confusing the direction, mixing up standard deviations with standard errors, or drastically miscalculating the z-score. In a normal distribution, about 16% of values fall more than 1 standard deviation above the mean, so this reading is elevated but not rare.

This problem involves interpreting a heart rate value in a normal distribution. Given μ = 60 bpm and σ = 8 bpm, we calculate the z-score for x = 44 bpm: z = (44 - 60)/8 = -16/8 = -2. A heart rate of 44 bpm is 2 standard deviations below the mean, making it relatively low for runners. The distractors include confusing the direction (above vs. below), mixing up standard deviations with standard errors, or miscalculating how many standard deviations away the value is. Values 2 standard deviations below the mean are relatively unusual, occurring in only about 2.5% of the population.

This question tests understanding of intervals in normal distributions. With μ = 150 g and σ = 10 g, we analyze the interval from 130 g to 170 g. First, standardize the endpoints: (130 - 150)/10 = -20/10 = -2 and (170 - 150)/10 = 20/10 = 2. The interval spans from 2 standard deviations below to 2 standard deviations above the mean. According to the empirical rule, approximately 95% of values in a normal distribution fall within 2 standard deviations of the mean. This means "almost all" apples (about 95%) have weights in this range. The key insight is recognizing that μ ± 2σ captures approximately 95% of the distribution, while μ ± 1σ captures only about 68%. Understanding these benchmarks helps interpret what proportion of the population falls within various intervals.

This problem requires calculating how many standard deviations a lightbulb lifetime is from the mean. With μ = 1200 hours and σ = 150 hours, the z-score for x = 900 hours is: z = (900 - 1200)/150 = -300/150 = -2. A lifetime of 900 hours is 2 standard deviations below the mean, making it shorter than typical and relatively uncommon. The distractors include reversing the interpretation (longer instead of shorter), confusing standard deviations with standard errors, or miscalculating the z-score. Values 2 standard deviations below the mean occur in only about 2.5% of cases, making this bulb's lifetime relatively unusual.

This problem involves interpreting an interval in a normal distribution of bearing diameters. With μ = 10.00 mm and σ = 0.04 mm, the interval 9.92 to 10.08 mm extends from (9.92-10.00)/0.04 = -2 to (10.08-10.00)/0.04 = +2 standard deviations from the mean. According to the empirical rule, approximately 95% of values fall within 2 standard deviations of the mean, which means most (but not all) bearings. The distractors incorrectly calculate the number of standard deviations, confuse standard deviation with standard error, or misinterpret the interval's position. The interval is symmetric around the mean, not shifted below it.

This question tests understanding of interpreting values in a normal distribution using z-scores. Given μ = 70 inches and σ = 3 inches, we need to find how many standard deviations 76 inches is from the mean. The z-score is (76 - 70)/3 = 6/3 = 2, meaning 76 inches is 2 standard deviations above the mean. The common distractor is confusing "standard deviation" with "standard error" - standard error relates to sampling distributions, not individual values. In a normal distribution, about 95% of values fall within 2 standard deviations of the mean, so a value 2 standard deviations away is relatively unusual, occurring in only about 2.5% of the population.

This question tests understanding of extreme values in a normal distribution. With μ = 300 ms and σ = 40 ms, we need to determine where 220 ms falls. Calculate the standardized value: (220 - 300)/40 = -80/40 = -2. This means 220 ms is 2 standard deviations below the mean. In a normal distribution, values that are 2 or more standard deviations from the mean are considered unusual, occurring in only about 2.5% of cases in each tail. Since this is 2σ below the mean, it represents an unusually fast reaction time. In the context of video games, lower reaction times are better, so this would be an exceptionally good performance. Understanding that values beyond 2σ from the mean are rare helps identify which outcomes are typical versus exceptional.