What z score in a normal distribution has 33% of all score above it?
Answer: A z score which has 33% of all scores above it, will have 67% of all scores below it.
To find the required z score, we need to find the z value corresponding to probability 0.67.
Using the standard normal table, we have:
Therefore, the z score = 0.44 has 33% of all score above it.
To solve for the confidence interval for the true average
percentage elongation, we use the z statistic. The formula for confidence
interval is given as:
Confidence interval = x ± z σ / sqrt (n)
where,
x = the sample mean = 8.63
σ = sample standard deviation = 0.79
n = number of samples = 56
From the standard distribution tables, the value of z at
95% confidence interval is:
z = 1.96
Therefore substituting the known values into the
equation:
Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)
Confidence interval = 8.63 ± 0.207
Confidence interval = 8.42, 8.84
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Step-by-step explanation:
A
-----98 with any quantity of any digits before the 9 .
I believe Hector is correct
Explanation: for every 1 hour the income is $10, therefore working 32 hours will give you $320 as Hector stated, while working 18 hours will not give you $160 as Hannah stated, it will give you $180
