What is the probability of having an odd number in a single toss of a fair die
2 answers:
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Answer:
92.9997<<99.5203
Step-by-step explanation:
Using the formula for calculating the confidence interval expressed as:
CI = xbar ± Z * S/√n where;
xbar is the sample mean
Z is the z-score at 90% confidence interval
S is the sample standard deviation
n is the sample size
Given parameters
xbar = 96.52
Z at 90% CI = 1.645
S = 10.70.
n = 25
Required
90% confidence interval for the population mean using the sample data.
Substituting the given parameters into the formula, we will have;
CI = 96.52 ± (1.645 * 10.70/√25)
CI = 96.52 ± (1.645 * 10.70/5)
CI = 96.52 ± (1.645 * 2.14)
CI = 96.52 ± (3.5203)
CI = (96.52-3.5203, 96.52+3.5203)
CI = (92.9997, 99.5203)
<em>Hence a 90% confidence interval for the population mean using this sample data is 92.9997<</em><em><99.5203</em>