Answer :
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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(score greater than 615)
P(x > 615)
[tex]P( x > 615) = P( z > \displaystyle\frac{615 - 515}{100}) = P(z > 1)[/tex]
[tex]= 1 - P(z \leq 1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 615) = 1 - 0.8413 = 0.1587 = 15.87\%[/tex]
b) b) P(score greater than 715)
[tex]P(x > 715) = P(z > \displaystyle\frac{715-515}{100}) = P(z > 2)\\\\P( z > 2) = 1 - P(z \leq 2)[/tex]
Calculating the value from the standard normal table we have,
[tex]1 - 0.977 = 0.023 = 2.3\%\\P( x > 715) = 2.3\%[/tex]
c) P(score between 415 and 515)
[tex]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\%[/tex]
[tex]P(415 \leq x \leq 515) = 34.1\%[/tex]
d) P(score between 315 and 615)
[tex]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\%[/tex]
[tex]P(315 \leq x \leq 615) = 81.8\%[/tex]