Answer:
it has to be 60
Step-by-step explanation:
4 x 15 = 60
The missing part of the question is highlighted in bold format
The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age at first startup for 10% of entrepreneurs was 30 years of age or more.
Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. If required, round your answers to four decimal places. np = n(1-p) = E(p) = σ(p) = (b) Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. If required, round your answers to four decimal places.
Answer:
(a)
np = 180
n(1-p) = 20
E(p) = p = 0.9
σ(p) = 0.0212
(b)
np = 20
n(1 - p) = 180
E(p) = p = 0.1
σ(p) = 0.0212
Step-by-step explanation:
From the given information:
Let consider p to be the sample proportion of entrepreneurs whose first startup was at 29 years of age or less
So;
Given that :
p = 90% i.e p = 0.9
sample size n = 200
Then;
np = 200 × 0.9 = 180
n(1-p) = 200 ( 1 - 0.9)
= 200 (0.1)
= 20
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The expected value of the sampling distribution of p = E(p) = p = 0.9
Variance ![\sigma^2=\dfrac{p(1-p)}{n}](https://tex.z-dn.net/?f=%5Csigma%5E2%3D%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D)
![=\dfrac{0.9(1-0.9)}{200}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B0.9%281-0.9%29%7D%7B200%7D)
![=\dfrac{0.9(0.1)}{200}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B0.9%280.1%29%7D%7B200%7D)
![\mathbf{=4.5*10^{-4}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D4.5%2A10%5E%7B-4%7D%7D)
The standard error of σ(p) = ![\sqrt{\sigma ^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Csigma%20%5E2%7D)
![\mathbf{= \sqrt{4.5*10^{-4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20%5Csqrt%7B4.5%2A10%5E%7B-4%7D%7D%7D)
= 0.0212
(b)
Here ;
p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more
p = 10% i.e p = 0.1
sample size n = 200
Then;
np = 200 × 0.1 = 20
n(1 - p) = 200 (1 - 0.1 ) = 180
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The Expected value of the sampling distribution of p = E(p) = p = 0.1
Variance ![\sigma^2=\dfrac{p(1-p)}{n}](https://tex.z-dn.net/?f=%5Csigma%5E2%3D%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D)
![=\dfrac{0.1(1-0.1)}{200}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B0.1%281-0.1%29%7D%7B200%7D)
![=\dfrac{0.1(0.9)}{200}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B0.1%280.9%29%7D%7B200%7D)
![\mathbf{=4.5*10^{-4}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D4.5%2A10%5E%7B-4%7D%7D)
The standard error of σ(p) = ![\sqrt{\sigma ^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Csigma%20%5E2%7D)
![\mathbf{= \sqrt{4.5*10^{-4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20%5Csqrt%7B4.5%2A10%5E%7B-4%7D%7D%7D)
= 0.0212
Answer:
Yes she can.
Step-by-step explanation:
Val can make 6 slides in an hour.
1/6 * 60 minutes = 10 minutes. per slide
10 20 30 40 50 60 gives you six sldes, 1 every ten minutes.
=====================
In one hour, she can produce 6 slides.
In two hours, she can produce 12 slides (2 * 6 = 12)
Since she only needs 1 hour and 40 minutes to produce 10 slides, she is under her time limit of 2 hours.