Question

The mean score of a college entrance test is 500; the standard deviation is 75. the scores are normally distributed. what percent of the students scored below 320?

Answer 1

To find the percentage of students who scored below 320 on the college entrance test, calculate the z-score and use a standard normal distribution table. The percentage of students who scored below 320 is approximately 99.18%.

Step-by-step explanation:

To find the percentage of students who scored below 320 on the college entrance test, we need to calculate the z-score and then use a standard normal distribution table.

1. First, calculate the z-score using the formula: z = (x - μ) / σ, where x is the score and μ is the mean score, and σ is the standard deviation.
2. Next, consult a standard normal distribution table to find the percentile associated with the z-score.
3. Finally, subtract this percentile from 100 to find the percentage of students who scored below 320.

Using the given information, the z-score for a score of 320 is z = (320 - 500) / 75 = -2.4.

Looking up the percentile associated with a z-score of -2.4 in the standard normal distribution table, we find that it is approximately 0.0082.

Therefore, the percentage of students who scored below 320 is approximately 100 - (0.0082 * 100) = 99.18%.

Answer 2

To find the percentage of students who scored below 320 we proceed as follows:
σ=75
μ=500
the z-score of this information is given by:
z=(x-μ)/σ
thus the z-score when x=320 will be:
z=(320-500)/75
z=-2.4
thus:
P(x<320)=P(z=-2.4)=0.0082~0.82%
Thus:
Answer: 0.82%