I believe it is 5 I have the work in the picture
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%
Answer:
Step-by-step explanation:
(grey)
(green)
They are similar, the green is just double in size so a ratio of 1:2
Not sure what you're supposed to be doing here, but if you're supposed to factor...
-12*-10 = 120
-12+(-10) = -22
Thus,
(c - 12)(c - 10) = 0
c - 12 = 0
c = 12
or
c - 10 = 0
c = 10
Answer:
The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
Step-by-step explanation:
The null hypothesis is:
The alternate hypotesis is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.
A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position.
This means that
Assume the population standard deviation is 6.2 weeks.
This means that
Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
We have to find the pvalue of Z, looking at the z-table, when . It if is lower than 0.05, it provides evidence.
has a pvalue of 0.0113 < 0.05.
The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance