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Juli2301 [7.4K]
3 years ago
8

The national average for the math portion of the College Board’s Scholastic Aptitude Test

Mathematics
1 answer:
Alex73 [517]3 years ago
4 0

Answer:

a) 0.1587

b) 0.023

c) 0.341

d) 0.818

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 515

Standard Deviation, σ = 100

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(score greater than 615)

P(x > 615)

P( x > 615) = P( z > \displaystyle\frac{615 - 515}{100}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 615) = 1 - 0.8413 = 0.1587 = 15.87\%

b) b) P(score greater than 715)

P(x > 715) = P(z > \displaystyle\frac{715-515}{100}) = P(z > 2)\\\\P( z > 2) = 1 - P(z \leq 2)

Calculating the value from the standard normal table we have,

1 - 0.977 = 0.023 = 2.3\%\\P( x > 715) = 2.3\%

c) P(score between 415 and 515)

P(415 \leq x \leq 515) = P(\displaystyle\frac{415 - 515}{100} \leq z \leq \displaystyle\frac{515-515}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%

P(415 \leq x \leq 515) = 34.1\%

d) P(score between 315 and 615)

P(315 \leq x \leq 615) = P(\displaystyle\frac{315 - 515}{100} \leq z \leq \displaystyle\frac{615-515}{100}) = P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.841 - 0.023 = 0.818 = 81.8\%

P(315 \leq x \leq 615) = 81.8\%

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