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Sedaia [141]
3 years ago
14

The number of calories burned at the gym is normally distributed with a mean of 425 and a standard deviation of 51. Find the Z-s

core for each data value.
a.)268 b.)512 c.)450
Mathematics
1 answer:
Galina-37 [17]3 years ago
6 0

Answer:

(a) -3.708

(b) 1.706

(c) 0.490

Step-by-step explanation:

The z-score of normal distribution is given as:

z=\frac{x-\mu}{\sigma}

(a)

Given:

Score is, x=268

Mean value is, \mu =425

Standard deviation is, \sigma = 51

z-score is, z=\frac{268-425}{51}=\frac{-157}{51}=-3.078

(b)

Given:

Score is, x=512

Mean value is, \mu =425

Standard deviation is, \sigma = 51

z-score is, z=\frac{512-425}{51}=\frac{87}{51}=1.706

(c)

Given:

Score is, x=450

Mean value is, \mu =425

Standard deviation is, \sigma = 51

z-score is, z=\frac{450-425}{51}=\frac{25}{51}=0.490

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Suppose that at a state college, a random sample of 41 students is drawn, and each of the 41 students in the sample is asked to
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The  confidence interval for 90% confidence would be narrower than the 95% confidence

Step-by-step explanation:

From the question we are told that

  The  sample size is n = 41

   

For a 95% confidence the level of significance is  \alpha  = [100 - 95]\% =  0.05 and

the critical value  of  \frac{\alpha }{2}  is   Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }=  1.96

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Now the margin of error is mathematically represented as

         E =  Z_{\frac{\alpha }{2} } *  \frac{s}{\sqrt{n} }

given that other values are constant and only Z_{\frac{\alpha }{2} } is varying we have that

         E\ \  \alpha \ \   Z_{\frac{\alpha }{2} }

Hence for  reducing confidence level the margin of error will be reducing

  The  confidence interval is mathematically represented as

        \= x  - E  <  \mu <  \= x  + E

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value  reduces, the margin of error  reduces and the confidence interval becomes narrower

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