The probability that the mean clock life would differ from the population mean by greater than 12.5 years is 98.30%.
Given mean of 14 years, variance of 25 and sample size is 50.
We have to calculate the probability that the mean clock life would differ from the population mean by greater than 1.5 years.
μ=14,
σ=
=5
n=50
s orσ =5/
=0.7071.
This is 1 subtracted by the p value of z when X=12.5.
So,
z=X-μ/σ
=12.5-14/0.7071
=-2.12
P value=0.0170
1-0.0170=0.9830
=98.30%
Hence the probability that the mean clock life would differ from the population mean by greater than 1.5 years is 98.30%.
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There is a mistake in question and correct question is as under:
What is the probability that the mean clock life would differ from the population mean by greater than 12.5 years?
The correct labels are: Mean = C, Median = B and Mode = A
<h3>How to determine the labels?</h3>
The image that completes the question is added as an attachment
From the image, we can see that the graph is skewed left
In a left skewed distribution,
- mean is less than the median
- median is less than the mode
This means that:
Mean is the smallest and mode is the highest
Hence, the correct labels are:
Mean = C, Median = B and Mode = A
Read more about skewness at:
brainly.com/question/9329803
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Answer: 15
Step-by-step explanation:
Average is sum of scores/ number of scores, from the question, average is given as 15.
Let the missing score be x
Equation for the questionis represented as: 12 + 18 + x /3 = 15
30 + x /3 = 15
Cross multiply
30 + x = 15×3
30 + x = 45
Subtract 30 from both sides
30 + x -30 = 45 -30
x = 15.
Check:
12 + 18 + 15/3 = 15
45/3 = 15
I hope this is clear, please mark as brainliest answer
Answer:
D
Step-by-step explanation:
