Answer:
The probability is P(500 < X < 700 ) = 0.60044
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 544[/tex]
The standard deviation is [tex]\sigma = 103[/tex]
Generally the percentage of applicant that scored between 500 and 700 is mathematically represented as
[tex]P(500 < X < 700 ) = P(\frac{500 - 544}{103} < \frac{X - \mu }{\sigma } < \frac{700 - 544}{103} )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
[tex]P(500 < X < 700 ) = P(-0.4272 < Z < 1.5146 )[/tex]
=> [tex]P(500 < X < 700 ) = P( Z< 1.5146 ) - P ( Z < -0.4272 )[/tex]
Generally from the z-table, the area under the normal curve to the left corresponding to 1.5146 and -0.4272 is
P( Z< 1.5146 ) = 0.93506
P ( Z < -0.4272 ) = 0.33462
So
P(500 < X < 700 ) = 0.93506 - 0.33462
=> P(500 < X < 700 ) = 0.60044
