Answer:
387420489
Step-by-step explanation:
hope this helped
Answer:
A normal model is a good fit for the sampling distribution.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
The standard deviation of this sampling distribution of sample proportion is:
![\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The information provided is:
<em>N</em> = 675
<em>X</em>₁ = bodies with low vitamin-D levels had weak bones
<em>n</em>₁ = 82
<em>p</em>₁ = 0.085
<em>X</em>₂ = bodies with regular vitamin-D levels had weak bones
<em>n</em>₂ = 593
<em>p</em>₂ = 0.01
Both the sample sizes are large enough, i.e. <em>n</em>₁ = 82 > 30 and <em>n</em>₂ = 593 > 30.
So, the central limit theorem can be applied to approximate the sampling distribution of sample proportions by the Normal distribution.
Thus, a normal model is a good fit for the sampling distribution.
This answer would be .25, otherwise known as 25 cents
P(TTTTT)=P(T)^5 and P(T)=1/2 so
P(TTTTT)=(1/2)^5
P(TTTTT)=1/32