Answer:
a) 81.5%
b) 95%
c) 75%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 266 days
Standard Deviation, σ = 15 days
We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.
Formula:

a) P(between 236 and 281 days)

b) a) P(last between 236 and 296)

c) If the data is not normally distributed.
Then, according to Chebyshev's theorem, at least
data lies within k standard deviation of mean.
For k = 2

Atleast 75% of data lies within two standard deviation for a non normal data.
Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.
Answer:
a) ![[-0.134,0.034]](https://tex.z-dn.net/?f=%5B-0.134%2C0.034%5D)
b) We are uncertain
c) It will change significantly
Step-by-step explanation:
a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.
Since we assume that the variances are equal, we use the pooled variance given as
,
where
.
The mean difference
.
The confidence interval is

![= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]](https://tex.z-dn.net/?f=%3D%20-0.05%5Cpm%201.995%20%5Ctimes%200.042%20%3D%20-0.05%20%5Cpm%200.084%20%3D%20%5B-0.134%2C0.034%5D)
b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.
c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.
I would start at 2000 because of 400/.2 is 2000 idk sorry
Answer:
it 180 degree try to add them all the subtract with 180 and that equals x
Step-by-step explanation:
5.00 is the answer! It is very close to 5 therefore it would immediately go to 5!