Respuesta :
Answer:
a) [tex]P(X>0.5)=P(\frac{X-\mu}{\sigma}>\frac{0.5-\mu}{\sigma})=P(Z>\frac{0.5-0.4}{0.05})=P(z>2)[/tex]
[tex]P(z>2)=1-P(z<2)[/tex]
[tex]P(Z>2) = 1-P(Z<2)= 1- 0.97725=0.02275[/tex]
b)[tex]P(0.4<X<0.5)=P(\frac{0.4-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{0.5-\mu}{\sigma})=P(\frac{0.4-0.4}{0.05}<Z<\frac{0.5-0.4}{0.05})=P(0<z<2)[/tex]
[tex]P(0<z<2)=P(z<2)-P(z<0)[/tex]
[tex]P(0<z<2)=P(z<2)-P(z<0)=0.97725-0.5=0.47725[/tex]
c) [tex]a=0.4 +1.28*0.05=0.464[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 0.464.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the reaction time of a driver to visual stimulus of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(0.4,0.05)[/tex]
Where [tex]\mu=0.4[/tex] and [tex]\sigma=0.05[/tex]
We are interested on this probability
[tex]P(X>0.5)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>0.5)=P(\frac{X-\mu}{\sigma}>\frac{0.5-\mu}{\sigma})=P(Z>\frac{0.5-0.4}{0.05})=P(z>2)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>2)=1-P(z<2)[/tex]
And using the normal standard table or excel we have this:
[tex]P(Z>2) = 1-P(Z<2)= 1- 0.97725=0.02275[/tex]
Part b
[tex]P(0.4<X<0.5)=P(\frac{0.4-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{0.5-\mu}{\sigma})=P(\frac{0.4-0.4}{0.05}<Z<\frac{0.5-0.4}{0.05})=P(0<z<2)[/tex]
And we can find this probability on this way:
[tex]P(0<z<2)=P(z<2)-P(z<0)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(0<z<2)=P(z<2)-P(z<0)=0.97725-0.5=0.47725[/tex]
Part c
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.10[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.10 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28=\frac{a-0.4}{0.05}[/tex]
And if we solve for a we got
[tex]a=0.4 +1.28*0.05=0.464[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 0.464.
Answer:
a) 0.0228
b) 0.4772
c) 0.336
Step-by-step explanation:
Mean(μ) = 0.4 seconds
Standard deviation (σ) = 0.05 seconds
From normal distribution,
Z= (x - μ) / σ
a) P(x > 0.5)
Let x be the random variable for the required seconds
When x= 0.5
Z = (0.5 - 0.4)/0.05
Z = 2
From the normal distribution table, 2= 0.4772
φ(Z) = 0.4772
Recall that when Z is positive,
P(x >a) = 0.5 - φ(Z)
P(x > 0.5) = 0.5 - 0.4772
= 0.0228
b) For x = 0.4
Z= (x - μ) / σ
= (0.4 - 0.4) / 0.05
= 0
For x= 0.5
Z= (x - μ) / σ
= (0.5 - 0.4) / 0.05
= 2
From the table, P(0.4 < x < 0.5) = P(0 < Z < 2)
So we have
P(Z < 2) - P(Z<0)
From the table, 2 = 0.4772 and 0 = 0
We then have
0.4772 - 0
= 0.4772
c) we are looking for x such that 90% of the values lie above it or 10% of the value lie below it.
From the table , 10% probability gives a z value of -1.28
x = μ + Zσ
x = 0.4 + (-1.28*0.05)
x = 0.4 - (1.28*0.05)
= 0.336
