Answer:
the confidence interval for the true weight of the specimen is;
4.1593 ≤ μ ≤ 4.1611
Step-by-step explanation:
We are given;
Standard deviation; σ = 0.001
Sample size; n = 8
Average weight; x¯ = 4.1602
We are given a 99% confidence interval and the z-value at this interval is 2.576
Formula for confidence interval is;
CI = x¯ ± (zσ/√n)
Plugging in the relevant values, we have;
CI = 4.1602 ± (2.576 × 0.001/√8)
CI = 4.1602 ± 0.000911
CI = (4.1602 - 0.000911), (4.1602 + 0.000911)
CI ≈ (4.1593, 4.1611)
Thus, the confidence interval for the true weight will be expressed as;
4.1593 ≤ μ ≤ 4.1611
Where μ represents the true weight
Answer:
456
Step-by-step explanation:
Let X be the SATscore scored by the students
Given that X is normal (1000,200)
By converting into standard normal variate we can say that
is N(0,1)
To find the top 10% we consider the 90th percentile for z score
Z 90th percentile = 1.28
i.e. only students who scored 456 or above only should be considered.
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i hope this is the answer im glad to help
Answer:
Explanation:
simplify the following
subtract 1 from both sides
simplify the following
multiply both sides by 1/3
simplify the following
