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:

$P(X<230)=P(\frac{X-\mu}{\sigma}<\frac{230-210}{\sqrt{32}})=P(Z<3.54)=0.9998$

*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:

$P(180<X<245)=P(\frac{180-210}{\sqrt{32}} <\frac{X-\mu}{\sigma} <\frac{245-210}{\sqrt{32}})\\=P(-5.30<Z<6.19)\\\approx 1$

*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:

$P(X>190)=P(\frac{X-\mu}{\sigma}>\frac{190-210}{\sqrt{32}})=P(Z>-3.54)=P(Z<3.54)=0.9998$

*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.

$P(X<c)=0.0344\\P(\frac{X-\mu}{\sigma}<\frac{c-210}{\sqrt{32}} )=0.0344\\P(Z<z)=0.0344$

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.

$P(X>c)=0.7486\\P(\frac{X-\mu}{\sigma}>\frac{c-210}{\sqrt{32}} )=0.7486\\P(Z>z)=0.7486\\P(Z<z)=1-0.7486=0.2514$

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.

Answer 2

Using the normal distribution, it is found that:

a) P(X < 230) = 0.734.

b) P(180 < X < 245) = 0.6885.

c) P( X >190) = 0.734.

d) X = 151.76.

e) X = 188.56.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is of $\mu = 210$.

• The standard deviation is of $\sigma = 32$.

Item a:

This probability is the p-value of Z when X = 230, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{230 - 210}{32}$

$Z = 0.625$

$Z = 0.625$ has a p-value of 0.734.

Hence:

P(X < 230) = 0.734.

Item b:

This probability is the p-value of Z when X = 245 subtracted by the p-value of Z when X = 180, hence:

X = 245

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{245 - 210}{32}$

$Z = 1.09$

$Z = 1.09$ has a p-value of 0.8621.

X = 180

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{180 - 210}{32}$

$Z = -0.94$

$Z = -0.94$ has a p-value of 0.1736.

0.8621 - 0.1736 = 0.6885.

Then:

P(180 < X < 245) = 0.6885.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 190, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{190 - 210}{32}$

$Z = -0.625$

$Z = -0.625$ has a p-value of 0.266.

1 - 0.266 = 0.734.

Hence:

P( X >190) = 0.734.

Item d:

This is X = c when Z has a p-value of 0.0344, hence X when Z = -1.82.

$Z = \frac{X - \mu}{\sigma}$

$-1.82 = \frac{X - 210}{32}$

$X - 210 = -1.82(32)$

$X = 151.76$

Item e:

This is X when Z has a p-value of 1 - 0.7486 = 0.2514, hence X when Z = -0.67.

$Z = \frac{X - \mu}{\sigma}$

$-0.67 = \frac{X - 210}{32}$

$X - 210 = -0.67(32)$

$X = 188.56$

More can be learned about the normal distribution at https://brainly.com/question/24663213