Respuesta :
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:
[tex]P(X<230)=P(\frac{X-\mu}{\sigma}<\frac{230-210}{\sqrt{32}})=P(Z<3.54)=0.9998[/tex]
*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:
[tex]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[/tex]
*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:
[tex]P(X>190)=P(\frac{X-\mu}{\sigma}>\frac{190-210}{\sqrt{32}})=P(Z>-3.54)=P(Z<3.54)=0.9998[/tex]
*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.
[tex]P(X<c)=0.0344\\P(\frac{X-\mu}{\sigma}<\frac{c-210}{\sqrt{32}} )=0.0344\\P(Z<z)=0.0344[/tex]
The value of z for which P (Z < z) = 0.0344 is -1.82.
Compute the value of c as follows:
[tex]-1.82=\frac{c-210}{\sqrt{32}} \\c=210-1.82\times/\sqrt{32}\\=210-10.30\\=199.70[/tex]
Thus, the value of c is 199.7.
(e)
It is provided that P (X > c) = 0.7486.
[tex]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[/tex]
The value of z for which P (Z < z) = 0.2514 is 0.67.
Compute the value of c as follows:
[tex]0.67=\frac{c-210}{\sqrt{32}} \\c=210+0.67\times/\sqrt{32}\\=210+3.79\\=213.79[/tex]
Thus, the value of c is 213.79.
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 210[/tex].
- The standard deviation is of [tex]\sigma = 32[/tex].
Item a:
This probability is the p-value of Z when X = 230, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{230 - 210}{32}[/tex]
[tex]Z = 0.625[/tex]
[tex]Z = 0.625[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{245 - 210}{32}[/tex]
[tex]Z = 1.09[/tex]
[tex]Z = 1.09[/tex] has a p-value of 0.8621.
X = 180
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{180 - 210}{32}[/tex]
[tex]Z = -0.94[/tex]
[tex]Z = -0.94[/tex] 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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{190 - 210}{32}[/tex]
[tex]Z = -0.625[/tex]
[tex]Z = -0.625[/tex] 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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.82 = \frac{X - 210}{32}[/tex]
[tex]X - 210 = -1.82(32)[/tex]
[tex]X = 151.76[/tex]
Item e:
This is X when Z has a p-value of 1 - 0.7486 = 0.2514, hence X when Z = -0.67.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.67 = \frac{X - 210}{32}[/tex]
[tex]X - 210 = -0.67(32)[/tex]
[tex]X = 188.56[/tex]
More can be learned about the normal distribution at https://brainly.com/question/24663213
