Respuesta :
Answer:
a) No. It is not normal.
b) The probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people is 0.104
Step-by-step explanation:
(a) Could the exact distribution of the count be Normal?
The exact distribution of the number of people in each car entering a freeway at a suburban interchange is not normal. Because the count is discrete and can assume values bigger or equal to one.
(b) The probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people.
The probability we seek is the cars carrying people with mean more than [tex]\frac{1075}{700}=1.5357[/tex]
That is P(z>z*) where z* is the z-score of 1.5357.
z* can be calculated using the equation:
z*=[tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] where
- X is the mean value wee seek for its z-score (1.5357)
- M is the average count of people entering a freeway at a suburban interchange. (1.5)
- s is the standard deviation of the count (0.75)
- N is the sample size (700)
Thus z*=[tex]\frac{1.5357-1.5}{\frac{0.75}{\sqrt{700} } }[/tex] ≈ 1.26
We have P(z>1.26)=1-P(z≤1.26)= 1-0.896 = 0.104
Probability distribution is a statistical function that describe all the positive value and similar patter that a random variable can take in a known range. The exact distribution of the count could not be normal and the probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people is 1.26.
Given-
Standard deviation [tex]s[/tex] is given which is 0.75.
The sample size n is given in the question which is 700.
What is Probability distribution?
Probability distribution is a statistical function that describe all the positive value and similar pattern that a random variable can take in a known range.
- (a) Could the exact distribution of the count be Normal?-No the exact distribution of the count could not be normal. As the given count is distinct the value may be larger or equal to the one.
- b) the probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people-For 1075 people the 700 randomly selected. Hence the probability P must be more than the,
[tex]=\dfrac{1075}{700}= 1.5357[/tex]
Now the critical value of standard score [tex]z[/tex] to provide region of rejection can be calculated as,
[tex]z^*=\dfrac{X-M}{\dfrac{s}{\sqrt{n} } }[/tex]
Here standard deviation [tex]s[/tex] is given which is 0.75.
The mean value for z score [tex]X[/tex] is 1.5357 has been already calculated.
The sample size n is given in the question which is 700.Thus,
[tex]z^*=\dfrac{1.5357-1.5}{\dfrac{0.75}{\sqrt{700} } }[/tex]
[tex]z^*=1.26[/tex]
The p value for this z score in z table is 0.104.
Hence the exact distribution of the count could not be normal and the probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people is 1.26.
Learn more about the Probability distribution.
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