Solution: We are given:
![\mu=0.33, \sigma =0.7](https://tex.z-dn.net/?f=%5Cmu%3D0.33%2C%20%20%5Csigma%20%3D0.7)
Let
be the weight (oz) of laptop
We have to find ![P(x>32)](https://tex.z-dn.net/?f=P%28x%3E32%29)
To find the this probability, we need to find the z score value.
The z score is given below:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![=\frac{32-33}{0.7}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B32-33%7D%7B0.7%7D)
![=-1.43](https://tex.z-dn.net/?f=%3D-1.43)
Now, we have to find ![P(z>-1.43)](https://tex.z-dn.net/?f=P%28z%3E-1.43%29)
Using the standard normal table, we have:
![P(z>-1.43)=0.9236](https://tex.z-dn.net/?f=P%28z%3E-1.43%29%3D0.9236)
0.9236 or 92.36% of laptops are overweight
Answer:
The margin of error for the 99% confidence level for this sample is ±2.23.
None of the given figures is close to the answer:
E. None of the above
Step-by-step explanation:
margin of error (ME) around the mean can be calculated using the formula
ME=
where
- z is the corresponding statistic of the 99% confidence level (2.576)
- s is the population IQ standard deviation (15)
- N is the sample size (300)
Using these numbers we get:
ME=
≈ 2.23
Answer:
Step-by-step explanation:
x = number of pushups he did
Answer:
20 x
Step-by-step explanation:
i hope that is it
Answer:
b -3,3 c 6,9 hope this helps