Employee C is paid an hourly rate of $13.25 per hour and works 2,080 hours per year. The employee was just promoted and received
2 answers:
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Answer:
<em>a)95% confidence intervals for the population mean of light bulbs in this batch</em>
(325.5 ,374.5)
b)
<em>The calculated value Z = 4 > 1.96 at 0.05 level of significance</em>
<em>Null hypothesis is rejected </em>
<em>The manufacturer has not right to take the average life of the light bulbs is 400 hours.</em>
Step-by-step explanation:
Given sample size n = 64
Given mean of the sample x⁻ = 350
Standard deviation of the Population σ = 100 hours
The tabulated value Z₀.₉₅ = 1.96
<em>95% confidence intervals for the population mean of light bulbs in this batch</em>
<em></em><em></em>
<em></em><em></em>
(325.5 ,374.5)
b)
<u><em>Explanation</em></u>:-
Given mean of the Population μ = 400
Given sample size n = 64
Given mean of the sample x⁻ = 350
Standard deviation of the Population σ = 100 hours
<u><em>Null hypothesis</em></u> : H₀:The manufacturer has right to take the average life of the light bulbs is 400 hours.
μ = 400
<u><em>Alternative Hypothesis: H₁:</em></u> μ ≠400
<u><em>The test statistic </em></u>
|Z| = |-4|
The tabulated value Z₀.₉₅ = 1.96
The calculated value Z = 4 > 1.96 at 0.05 level of significance
Null hypothesis is rejected.
<u><em>Conclusion:</em></u>-
The manufacturer has not right to take the average life of the light bulbs is 400 hours.
Answer:
(a) The probability of exactly three flaws in 150 m of cable is 0.21246
(b) The probability of at least two flaws in 100m of cable is 0.69155
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063
Step-by-step explanation:
A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by
for x = 0, 1, 2, ...
where , the mean number of successes.
(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is
The probability of exactly three flaws in 150 m of cable is
(b) The probability of at least two flaws in 100m of cable is,
we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is
The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is
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