Question

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The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 998 and a standard deviation of 202. The College of Westport includes a minimum score of 1100 as one of its requirements for admission. a) What percentage of females do not satisfy the requirement? b) If the requirement is changed to a score that is in the top 40%, what is the minimum required score?

Answer 1

Solution :

Given that ,

a) P(x < 1100)

= P[(x - $\mu$ ) / $\sigma$ < (1100 - 998) / 202 ]

= P(z < 0.5050)

Using z table,

= 0.6932

percentage = 69.32%

b) Using standard normal table,

P(Z > z) = 40%

= 1 - P(Z < z) = 0.40  

= P(Z < z) = 1 - 0.40

= P(Z < 0.2534 ) = 0.60  

z = 0.2534

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 0.2534 * 202 + 998

x = 1049.19

The minimum required score = 1049