The scores of 12th-grade students on the national assessment of educational progress year 2000 mathematics test have a distribution that is approximately normal with mean of 300 and standard deviation of 35. a. choose one 12th-grader at random. what is the probability that his or her score is higher than 300? higher than 335? b. now choose an srs of four 12th-graders. what is the probability that their mean score is higher than 300? higher than 335?

Step-by-step explanation:

The scores of 12th-grade students on the National Assessment of Educational Progress year 2000 mathematics test have a distribution that is approximately Normal with mean µ = 300 and standard deviation σ = 35.

(a)Choose one 12th-grader at random. What is the probability that his or her score is higher than 300? Higher than 335?

The mean is 300, so the probability of a higher score is about 0.5. P(>300) = 0.5. A score of 335 is one standard deviation above the mean, so by the 68 part of the 68-95-99.7 rule the probability of a higher score is half of 0.32, or 0.16. P(>335) = 0.1587.

(b) Now choose an SRS of four 12th-graders and calculate their mean score x. If you did this many times, what would be the mean and standard deviation of all the x-values?

The average score of n = 4 students has a mean of 300 (µ=300). The standard deviation of the sample would be σ/√n = 35/√4 = 35/2 = 17.5. (  = 17.5).

(c)What is the probability that the mean score for your SRS is higher than 300? Higher than 335?

The average score of n = 4 students has mean 300 and standard deviation of the sample = σ/√n = 35/√4 = 35/2 = 17.5. The probability is 17.5. The probability of an average score higher than 300 is still 0.5 because 335 is now two standard deviations above the mean, the 95 part of the 68-95-99.7 rule says that the probability of a higher average score is 0.025.

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

answer: the common difference is -0.10