Question

The mean of a normal probability distribution is 500; the standard deviation is 14. a. About 68% of the observations lie between what two values? Value 1 Value 2 b. About 95% of the observations lie between what two values? value a Value 1 Value 2 c. Practically all of the observations lie between what two values? Value 1 Value 2

Answer 1

Solution :

Given that ,

mean = $\mu$ = 500

standard deviation = $\sigma$ = 14

Using Empirical rule,

a) P( $\mu$ - $\sigma$ < x <  $\mu$ + $\sigma$ ) = 68%

= P( 500 - 14 < x < 500 + 14 ) = 68%

= P( 486 < x < 514 ) =68%

value 1 = 486

value 2 = 514

b) P( $\mu$ - 2$\sigma$ < x <  $\mu$ + 2$\sigma$ ) = 95%

= P( 500 - 2 * 14 < x < 500 + 2 * 14 ) = 95%

= P( 500 - 28 < x < 500 + 28 ) = 95%

=P( 472 < x < 528 ) = 95%

value 1 = 472

value 2 = 528

c) P( $\mu$ - 3$\sigma$ < x <  $\mu$ + 3$\sigma$ ) = 99.7%

= P( 500 - 3 * 14 < x < 500 + 3 * 14 ) = 99.7%

= P( 500 - 42 < x < 500 + 42 ) = 99.7%

=P( 458 < x < 542 ) = 99.7%

value 1 = 458

value 2 = 542