Please help explanation if possible
1 answer:
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The normal distribution curve for the problem is shown below
We need to standardise the value X=405.5 by using the formula
We now need to find the probability of z=0.32 by reading the z-table
Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:
from the z-table, the reading
gives 0.6255
hence,
The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%
We need to standardise the value X=405.5 by using the formula
We now need to find the probability of z=0.32 by reading the z-table
Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:
from the z-table, the reading
hence,
The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%
Answer:
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
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:
For this problem, we have that:
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is
We need a sample size of at least n, in which n is found M = 0.04.
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.