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
Answer:
The correct answer is;
A. 0.17
Step-by-step explanation:
Here are the provided details;
The average time taken for a cashier to handle an order, μ = 276 seconds
The deviation from the average, σ = 38 seconds
The z-score for an order processing time of x = 240 seconds can be calculated as follows;
Thus;
The resulting probability
P(z = -0.9474) = 0.17361
Hence, the estimated proportion of orders processed in under 240 seconds is roughly 0.17361 or 0.17 when rounded to two decimal places.
There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.
Answer:
In this case, the method of sampling employed by the poll
Explanation:
Convenience sampling, often referred to as grab, opportunity, or accidental sampling, is a non-probability sampling technique where the sample is collected from the part of the population that is easily accessible. This technique proves to be particularly useful for preliminary investigations.
This type of sampling makes selections based on ease of access, willingness of individuals to participate, and availability at specific times.
Consequently, this approach can lead to biased results and may not yield the expected outcomes.
They cannot possess the same number of horses; let me clarify. If you divide 21 horses among four individuals, you would perform 21/4, yielding 5.25, implying that fractional horses are unfeasible. Therefore, at least one individual must have 6 horses instead of 5. Possibly, this is what your instructor wants you to understand. For an even distribution, they could sell one horse, making it 20, so each would then have 5 horses. Alternatively, they might share the extra horse to rotate its usage.
