Question

Suppose X is a normal distribution with N(210, 32). Find the following: a. P( X < 230) b. P(180 < X < 245) c. P( X >190) d. Find c such that P( X < c) = 0.0344 e. Find c such that P( X > c) = 0.7486

Answer 1

Answer:

(a) The value of P (X < 230) is 0.9998.

(b) The value of P (180 < X < 245) is 1.

(c) The value of P (X > 190) is 0.9998.

(d) The value of c is 199.7.

(e) The value of c is 213.79.

Step-by-step explanation:

It is provided that random variable X follows a Normal distribution with parameters μ = 210 and σ = √32.

(a)

Compute the probability of the event X < 230 as follows:

$P(X$

*Use the z-table for the probability.

Thus, the value of P (X < 230) is 0.9998.

(b)

Compute the probability of the event 180 < X < 245 as follows:

$P(180$

*Use the z-table for the probability.

Thus, the value of P (180 < X < 245) is 1.

(c)

Compute the probability of the event X > 190 as follows:

$P(X>190)=P(\frac{X-\mu}{\sigma}>\frac{190-210}{\sqrt{32}})=P(Z>-3.54)=P(Z$

*Use the z-table for the probability.

Thus, the value of P (X > 190) is 0.9998.

(d)

It is provided that P (X < c) = 0.0344.

$P(X$

The value of z for which P (Z < z) = 0.0344 is -1.82.

Compute the value of c as follows:

$-1.82=\frac{c-210}{\sqrt{32}} \\c=210-1.82\times/\sqrt{32}\\=210-10.30\\=199.70$

Thus, the value of c is 199.7.

(e)

It is provided that P (X > c) = 0.7486.

$P(X>c)=0.7486\\P(\frac{X-\mu}{\sigma}>\frac{c-210}{\sqrt{32}} )=0.7486\\P(Z>z)=0.7486\\P(Z$

The value of z for which P (Z < z) = 0.2514 is 0.67.

Compute the value of c as follows:

$0.67=\frac{c-210}{\sqrt{32}} \\c=210+0.67\times/\sqrt{32}\\=210+3.79\\=213.79$

Thus, the value of c is 213.79.