Question

In: Statistics and Probability

Scores on the SAT for the class of 2007 were roughly normal with a mean of...

Scores on the SAT for the class of 2007 were roughly normal with a mean of 1511 and a standard deviation of 194. (a) What was the range of the middle 68% of SAT scores? -- (b) What was the range of the middle 95% of SAT scores? -- (c) How high must a student score to be in the top 2.5% of SAT scores?

Solutions

Expert Solution

Solution : -

Given that,

mean = = 1511

standard deviation = = 194

Middle 68%

= 1 - 68%

= 1 - 0.68 = 0.32

/2 = 0.16

1 - /2 = 1 - 0.16 = 0.84

Z/2 = Z0.16 = -0.994

Z 1 - /2 = Z0.84 = 0.994

Using z-score formula,

x = z * +

x = -0.994 * 194 + 1511

x = 1318.164

Using z-score formula,

x = z * +

x = 0.994 * 194 + 1511

x = 1703.836

Answer = Between 1318 and 1704

( b )

Middle 95%

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

1 - /2 = 1 - 0.025 = 0.975

Z/2 = Z0.025 = -1.96

Z1 - /2 = Z0.975 = 1.96

Using z-score formula,

x = z * +

x = -1.96 * 194 + 1511

x = 1130.76

Using z-score formula,

x = z * +

x = 1.96 * 194 + 1511

x = 1891.24

Answer = Between 1131 and 1891

( c )

The z - distribution of the 2.5% is ,

P(Z > z) = 2.5%

= 1 - P(Z < z ) = 0.025

= P(Z < ) = 1 - 0.025

= P(Z < z ) = 0.975

= P(Z < 1.960 ) = 0.975

z = 1.960

Using z-score formula,

x = z * +

x = 1.960 * 194 + 1511

x = 1891.24

Answer = x = 1891

orchestra answered 1 year ago

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