We can formulate the trajectory of the parabola using the vertex form equation: y = a (x – h)^2 + k. The coordinates for the vertex are at h and k, representing the peak height, thus h = 250 and k = 120. Consequently, the equation becomes y = a (x – 250)^2 + 120. At the starting point where x = 0 and y = 0, we find a: 0 = a (0 – 250)^2 + 120, which simplifies to 0 = a (62,500) + 120, leading to a = -0.00192. The complete equation is y = -0.00192 (x – 250)^2 + 120. To determine y when x = 400, we substitute: y = -0.00192 (400 - 250)^2 + 120, yielding y = 76.8 ft. Hence, the ball clears the tree by 76.8 ft – 60 ft = 16.8 ft.
Answer:
Answer and Explanation:
We have:
Population mean,
μ
=
3
,
000
hours
Population standard deviation,
σ
=
696
hours
Sample size,
n
=
36
1) The standard deviation for the sampling distribution:
σ
¯
x
=
σ
√
n
=
696
√
36
=
116
2) By the central limit theorem, the sampling distribution's expected value matches the population mean.
Thus:
The expected value of the sampling distribution equals the population mean,
μ
¯
x
=
μ
=
3
,
000
The standard deviation of the sampling distribution,
σ
¯
x
=
116
The sampling distribution of
¯
x
is roughly normal due to a sample size greater than
30
.
3) The likelihood that the average lifespan of the sample falls between
2670.56
and
2809.76
hours:
P
(
2670.56
<
x
<
2809.76
)
=
P
(
2670.56
−
3000
116
<
z
<
2809.76
−
3000
116
)
=
P
(
−
2.84
<
z
<
−
1.64
)
=
P
(
z
<
−
1.64
)
−
P
(
z
<
−
2.84
)
=
0.0482
In Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)
4) The probability of the average life in the sample exceeding
3219.24
hours:
P
(
x
>
3219.24
)
=
P
(
z
>
3219.24
−
3000
116
)
=
P
(
z
>
1.89
)
=
0.0294
In Excel: =NORMSDIST(-1.89)
5) The likelihood that the sample's average life is lower than
3180.96
hours:
P
(
x
<
3180.96
)
=
P
(
z
<
3180.96
−
3000
116
)
=
P
(
z
<
1.56
)
=
0.9406
Response:
The data shows skewness, with the minimum amount of crackers in a pack being 7
Detailed explanation:
Hello,
Firstly, the question lacks completeness due to missing information from the box plot, which I have provided to assist you in answering your inquiry.
Considering the details from the attached image, a symmetric distribution would be centered evenly, but that is not the case here.
The image indicates a positive skew, with the lowest count recorded as 7.
