Find the area to the right of the z-score 1.39 and to the left of the z-score 1.53 under the standard normal curve.1.2 1.3 1.4 0.00 0.01 0.02 0.03 0.8849 0.8869 0.8888 0.8907 0.9032 0.9049 0.9066 0.9082 0.9192 0.9207 0.9222 0.9236 0.9332 0.9345 0.9357 0.9370 0.9452 0.9463 0.9474 0.9484 0.9554 0.9564 0.9573 0.9582 0.04 0.05 0.06 0.070.08 0.09 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.5 1.6 1.7 Use the value(s) from the table above.

From the question we are told to obtain the area to the of the z-score 1.39 and to the left of the z-score 1.53, this is mathematically represented as

[tex]P( 1.39 < z < 1.53 ) = P( z < 1.53 ) - P( z < 1.39 )[/tex]

From the z table on the question the area under the normal curve to the left corresponding to 1.53 and 1.39 is

[tex]P( z < 1.53 ) = 0.9370[/tex]

and

[tex]P( z < 1.39 ) = 0.9177[/tex]

So

[tex]P( 1.39 < z < 1.53 ) = 0.9370 - 0.9177[/tex]

=> [tex]P( 1.39 < z < 1.53 ) = 0.0193[/tex]