5. Write the equation of the line graphed below.
1 answer:
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Answer:
Step-by-step explanation:
In the normal distribution curve, the mean is in the middle and each line to the left and to the right of that mean represent 1- and 1+ the standard deviation. If our mean is 400, then 400 + 50 = 450; 450 + 50 = 500; 500 + 50 = 550. Going from the mean to the left, we subtract the standard deviation and 400 - 50 = 350; 350 - 50 = 300; 300 - 50 = 250. We are interested in the range that falls between 350 and 450 as a percentage. That range represents the two middle sections, each containing 34% of the data. So the total percentage of response times is 68%. We are looking then for 68% of the 144 emergency response times in town. .68(144) = 97.92 or 98 emergencies that have response times of between 350 and 450 seconds.
f(x)=x^2+12x+26
Vertex form of f(x) = a(x-h)^2 +k
We change f(x) into vertex form using completing the square method
We take half of square of middle term
middle term is 12
Half of square of 12 is then 6^2 = 36
Add and subtract 36 to complete the square
The vertex form
Answer:
(a) The standard deviation of the amount spent is $3229.18.
(b) The probability that a household spends between $4000 and $6000 is 0.2283.
(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.
Step-by-step explanation:
We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.
(a) It is stated that 5% of American households spend less than $1000 for daily transportation.
Let X = <u><em>the amount spent on daily transportation</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = average annual amount American households spend on daily transportation = $6,312
= standard deviation
Now, 5% of American households spend less than $1000 on daily transportation means that;
P(X < $1,000) = 0.05
P( <
) = 0.05
P(Z < ) = 0.05
In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;
= 3229.18
So, the standard deviation of the amount spent is $3229.18.
(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)
P($4000 < X < $6000) = P(X < $6000) - P(X $4000)
P(X < $6000) = P( <
) = P(Z < -0.09) = 1 - P(Z
0.09)
= 1 - 0.5359 = 0.4641
P(X $4000) = P(
) = P(Z
-0.72) = 1 - P(Z < 0.72)
= 1 - 0.7642 = 0.2358
Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.
(c) The range of spending for 3% of households with the highest daily transportation cost is given by;
P(X > x) = 0.03 {where x is the required range}
P( >
) = 0.03
P(Z > ) = 0.03
In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;
x = $6312 + 6070.86 = $12382.86
So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.