Moshe has the container shown below on his desk to hold pencils and paper clips. 2 rectangular prisms. One has a length of 4 inc
2 answers:
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By assuming the standard deviation of population 2.2 the confidence interval is 8.67 toys,8.94 toys.
Given sample size of 1492 children,99% confidence interval , sample mean of 8.8, population standard deviation=2.2.
This type of problems can be solved through z test and in z test we have to first find the z score and then p value from normal distribution table.
First we have to find the value of α which can be calculated as under:
α=(1-0.99)/2=0.005
p=1-0.005=0.995
corresponding z value will be 2.575 for p=0.995 .
Margin of error=z*x/d
where x is mean and d is standard deviation.
M=2.575*2.2/
=0.14
So the lower value will be x-M
=8.8-0.14
=8.66
=8.67 ( after rounding)
The upper value will be x+M
=8.8+0.14
=8.94
Hence the confidence interval will be 8.67 toys and 8.94 toys.
Learn more about z test at brainly.com/question/14453510
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Answer:
The probability that none of the LED light bulbs are defective is 0.7374.
Step-by-step explanation:
The complete question is:
What is the probability that none of the LED light bulbs are defective?
Solution:
Let the random variable <em>X</em> represent the number of defective LED light bulbs.
The probability of a LED light bulb being defective is, P (X) = <em>p</em> = 0.03.
A random sample of <em>n</em> = 10 LED light bulbs is selected.
The event of a specific LED light bulb being defective is independent of the other bulbs.
The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 10 and <em>p</em> = 0.03.
The probability mass function of <em>X</em> is:
Compute the probability that none of the LED light bulbs are defective as follows:
Thus, the probability that none of the LED light bulbs are defective is 0.7374.