Answer:
(a) The value of P (X < 230) is 0.9998.
(b) The value of P (180 < <em>X</em> < 245) is 1.
(c) The value of P (X > 190) is 0.9998.
(d) The value of <em>c</em> is 199.7.
(e) The value of <em>c</em> is 213.79.
Step-by-step explanation:
It is provided that random variable <em>X</em> follows a Normal distribution with parameters <em>μ</em> = 210 and <em>σ</em> = √32.
(a)
Compute the probability of the event <em>X</em> < 230 as follows:
*Use the <em>z</em>-table for the probability.
Thus, the value of P (X < 230) is 0.9998.
(b)
Compute the probability of the event 180 < <em>X</em> < 245 as follows:
*Use the <em>z</em>-table for the probability.
Thus, the value of P (180 < <em>X</em> < 245) is 1.
(c)
Compute the probability of the event <em>X</em> > 190 as follows:
*Use the <em>z</em>-table for the probability.
Thus, the value of P (X > 190) is 0.9998.
(d)
It is provided that P (X < c) = 0.0344.
The value of <em>z</em> for which P (Z < z) = 0.0344 is -1.82.
Compute the value of <em>c</em> as follows:
Thus, the value of <em>c</em> is 199.7.
(e)
It is provided that P (X > c) = 0.7486.
The value of <em>z</em> for which P (Z < z) = 0.2514 is 0.67.
Compute the value of <em>c</em> as follows:
Thus, the value of <em>c</em> is 213.79.