(x+x/2),(y+y/2) you have two x values and two y values to substitute into the equation
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:
<u>(a) Could the exact distribution of the count be Normal?</u>
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 <em>discrete </em>and <em>can assume values bigger or equal to one</em>.
<u>(b) The probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people.</u>
The probability we seek is the cars carrying people with mean more than 
That is P(z>z*) where z* is the z-score of 1.5357.
z* can be calculated using the equation:
z*=
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*=
≈ 1.26
We have P(z>1.26)=1-P(z≤1.26)= 1-0.896 = 0.104
Answer: 0.878
Step-by-step explanation:
Given : P( Internet access) = 80.5%= 0.805
P(cable television) = 82% = 0.82
P( Internet access and cable television) = 74.7% =0.747
Formula: P(A or B)= P(A ) + P(B) - P(A and B)
So, P( Internet access or cable television) = P( Internet access) + P(cable television) -P( Internet access and cable television)
= 0.805+0.82-0.747
= 0.878
The probability that a randomly selected household in the survey had either Internet access or cable = 0.878