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:
2.98
Step-by-step explanation:
Answer:
a=4.2
Step-by-step explanation:
6.4-a=2.2
so 6.4-22=a
4.2=a
The 68 - 95 - 99.7 rule, gives the basis to solve this question.
It says that for a normal distribution 95% of the results are between the mean minus 2 standard deviations and the mean plus 2 standard deviations.
Here:
mean = 64.5 inches,
standard deviaton = 2.5 inches
mean - 2 standard deviations = 64.5 inches - 5 inches = 59.5 inches
mean + 2 standard deviations = 64.5 inches + 5 inches = 69.5 inches
Then, the answer is that 95% of women range approximately between 59.5 inches and 69.5 inches.
