Question

10. A measurement population has μ-: 15.00 and σ 2.00. (a) What percentage of measurements are expected to be between 12.00 and 18.00? Between 14.00 and 19.00? Between 10.00 and 11.00? (b) What interval around the mean will encompass 85% of the measurements? [Ans. (a) 86.64%, 66.88%, 1.65% (b) 15.00 +2.88)

Answer 1

10)

Solution :

Given that ,

mean = $\mu$ = 15

standard deviation = $\sigma$ = 2

(a) 1

P(12 < x < 18) = P((12 - 15) / 2) < (x - $\mu$ ) / $\sigma$ < (18 - 15) / 2) )

= P(-1.5 < z < 1.5)

= P(z < 1.5) - P(z < -1.5)

= 0.9332 - 0.0668

= 0.8664

86.64%

(a) 2

P(14 < x < 19) = P((14 - 15) / 2) < (x - $\mu$ ) / $\sigma$ < (19 - 15) / 2) )

= P(-0.5 < z < 2)

= P(z < 2) - P(z < -0.5)

= 0.9772 - 0.3085

= 0.6688

= 66.88%

(a) 3

P(10 < x < 11) = P((10 - 15) / 2) < (x - $\mu$ ) / $\sigma$ < (11 - 15) / 2) )

= P(-2.5 < z < -2)

= P(z < -2) - P(z < -2.5)

= 0.0228 - 0.0062

= 0.015

= 1.65%

(b)

z = $\pm$1.44

Using z-score formula,

x = z * $\sigma$ $\pm$  $\mu$

x = 15 $\pm$ 1.44 * 2 = 15 $\pm$ 2.88