A bricklayer needs to order 6300kg of building sand.
1 answer:
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We have been given that a colleague has been tutoring six students in 11th grade to prepare for the ACT. Student scores were as follows: 20, 18, 16, 15, 23, 20. We are asked to find the mean of the ACT scores.
We will use mean formula to solve our given problem.
Upon rounding to nearest whole number, we will get:
Therefore, the mean of the ACT scores is 19 and option 'c' is the correct choice.
Answer:
Step-by-step explanation:
Given that:
population mean = 10
standard deviation = 0.1
sample mean = 9.8 < x > 10.2
The z score can be computed as:
if x > 10.2
z = 2
If x < 9.8
z = -2
The p-value = P (z ≤ 2) + P (z ≥ 2)
The p-value = P (z ≤ 2) + ( 1 - P (z ≥ 2)
p-value = 0.022750 +(1 - 0.97725)
p-value = 0.022750 + 0.022750
p-value = 0.0455
Therefore; the probability of defectives = 4.55%
the probability of acceptable = 1 - the probability of defectives
the probability of acceptable = 1 - 0.0455
the probability of acceptable = 0.9545
the probability of acceptable = 95.45%
4.55% are defective or 95.45% is acceptable.
sampling distribution of proportions:
sample size n=1000
p = 0.0455
The z - score for this distribution at most 5% of the items is;
z = 0.6828
The p-value = P(z ≤ 0.6828)
From the z tables
p-value = 0.7526
Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526
The z - score for this distribution for at least 85% of the items is;
z = −15.86
p-value = P(z ≥ -15.86)
p-value = 1 - P(z < -15.86)
p-value = 1 - 0
p-value = 1
Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1