One plant from the garden is randomly selected. There is a 30% chance that the plant was purchased this year, and there is a 6%
chance that the plant was purchased this year and produces berries. What is the probability that the chosen plant does NOT produce berries, given that the plant was purchased this year? Write the probability as a percent. Round to the nearest tenth if needed. ____________%
2 answers:
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This problem is basically asking:
Which would mean
Answer:
The margin of error for the 90% confidence interval is of 0.038.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error is:
To this end we have obtained a random sample of 400 fruit flies. We find that 280 of the flies in the sample possess the gene.
This means that
90% confidence level
So , z is the value of Z that has a pvalue of
, so
.
Give the margin of error for the 90% confidence interval.
The margin of error for the 90% confidence interval is of 0.038.
Answer:
a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55
b)0.6004
c)19.607
Step-by-step explanation:
Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2
X ~ Poisson(A) where
a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar
50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55
So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55
b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment
Let X denotes the number of grams to be eaten before another fragment is detected.
c)The expected number of grams to be eaten before encountering the first fragments :
s