Question

section 1: multiple choice

1. many professional schools require applicants to take a standardized test. suppose that 1000 students take such a test. several weeks after the test, pete receives his score report: he got a 63, which placed him at the 73rd percentile. this means that:

a. pete did worse than about 63% of the test takers.
b. pete did worse than about 73% of the test takers.
c. pete did better than about 63% of the test takers.
d. pete did better than about 73% of the test takers.

1. the density curve shown models the distribution of a quantitative variable that is equally likely to take any value in the interval from 0 to 2. what percent of the observations lie between 0.5 and 1.2?

a. 25%
b. 35%
c. 50%
d. 70%

1. for the normal distribution shown, the standard deviation is closest to

a. 1
b. 2
c. 3
d. 6

1. scores on the act college entrance exam can be modeled using a normal distribution with mean 21 and standard deviation 5. waynes standardized score on the act was -0.6. what was waynes actual act score?

a. 3
b. 13
c. 16
d. 18

Explanation:

Step1: Define percentile

The nth percentile means a value is higher than n% of data points.

Step2: Apply to Pete's score

73rd percentile → better than ~73% of test takers.

Step1: Identify uniform distribution

Density curve is uniform on [0,2], so area = probability.

Step2: Calculate interval length

Length of [0.5,1.2] = 1.2 - 0.5 = 0.7.

Step3: Compute percentage

Percentage = 0.7 / 2 * 100% = 35%.

Step1: Use 68-95-99.7 rule

Normal distribution: ~95% data within μ±2σ.

Step2: Estimate range

From graph, ~95% data is between -6 and 6 (range 12). So 2σ = 6 → σ=3.

Step1: Recall z-score formula

z = (X - μ)/σ → X = μ + zσ.

Step2: Plug in values

μ=21, z=-0.6, σ=5 → X=21 + (-0.6)(5)=18.

Answer:

d. Pete did better than about 73% of the test takers.

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