The correct answer is A. According to the question's details, we're provided with key statistics on zucchini weights, which suggest that the average turns out to be typically 0.8 pounds, while the standard deviation is noted as 0.25 pounds. The probability of a randomly selected zucchini weighing between 0.55 pounds and 1.3 pounds can be mathematically expressed. Observing the provided normal distribution options, option A aligns with our specified weight range.
x y
1 290
2 280
3 270
4 260
5 250
6 240
7 230
8 220
9 210
10 200
11 90
12 180
13 170
14 160
15 150
16 140
17 130
18 120
19 110
20 100
21 90
22 80
23 70
24 60
25 50
26 40
27 30
28 20
29 10
30 0
Option D is indeed correct, as it ensures that the post's point is equidistant from the ground, maintaining a perpendicular angle at two points on the surface.
The P-value to evaluate the claim that the mean length of pencils produced in this factory equals 18.0 cm is 0.00736. Step-by-step explanation: In this case, a quality control specialist extracted a random sample of 45 pencils from the assembly line, which exhibited a mean length of 17.9 cm. With a known population standard deviation of 0.25 cm, we denote by the population mean length for pencils produced in the factory. Thus, Null Hypothesis: = 18.0 cm (indicating that the population mean length equals 18.0 cm). Alternate Hypothesis: ≠ 18.0 cm (suggesting different from 18.0 cm). We apply the one-sample z-test since the population standard deviation is known. The test statistic yields: T.S. ~ N(0,1), with the sample mean length 17.9 cm and population standard deviation 0.25 cm for a sample size of 45. Hence, the calculated test statistic is -2.68. The corresponding P-value is derived from P(Z < -2.68) = 1 - P(Z > 2.68), equating to 1- 0.99632 = 0.00368. For a two-tailed test, the resulting P-value computes to 2 * 0.00368 = 0.00736.
