Many states have programs for assessing the skills of students in various grades. The Indiana Statewide Testing for Educational Progress (ISTEP) is one such program. In a recent year, 76,531 tenth-grade Indiana students took the English/language arts exam. The mean score was 572 and the standard deviation was 51. Use the fact that the ISTEP scores are approximately Normal, N(572, 51). Find the proportion of students who have scores between 500 and 650.

Respuesta :

Answer:

P ( 500<X<650 ) = 0.8577

Step-by-step explanation:

Since μ=572 and σ=51 we have:

P ( 500<X<650 ) = P ( 500−572< X−μ<650−572 )

[tex]\RightarrowP ( \frac{500-572}{51} < \frac{x-\mu}{\sigma} < \frac{650-572}{51})[/tex]

⇒ P ( 500<X<650 ) = P ( −1.41<Z<1.53 )

Now, Using the standard normal table to conclude that:

P ( −1.41< Z <1.53 ) = 0.8577

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