The total number of tries = 10
The tries are { <span>110 111 100 000 101 111 100 000 011 010 }
</span><span />
Where: 0 representing heads and 1 representing tails
The tries which are heads came up more than once in 3 coin flips { 100 000 100 000 010 }
The number of the tries which are heads came up more than once in 3 coin flips = 5
∴ The probability of heads coming up more than once in 3 coin flips = 5/10 = 1/2
Answer:
4
Step-by-step explanation:
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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525 feet above the ground
