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Zigmanuir [339]
3 years ago
14

Suppose the scores of students on an exam are normally distributed with a mean of 244 and a standard deviation of 79. According

to the empirical rule, what percentage of students scored between 165 and 323 on the exam?
Mathematics
1 answer:
Ahat [919]3 years ago
3 0
244 - 165 = 79 . . . . one standard deviation
323 - 244 = 79 . . . . one standard deviation

The range of scores is ±1 standard deviation from the mean. The empirical rule says 
   68% of scores lie in that range.
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The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t
astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=\frac{ME*\sqrt{N}}{s } where

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  • s is the standard deviation of the population
  • N is the sample size

a.

z(p)=\frac{8*\sqrt{30}}{50 } ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=\frac{8*\sqrt{50}}{50 } ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=\frac{8*\sqrt{100}}{50 } ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥(\frac{z*s}{ME} )^2 where

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then N≥(\frac{1.96*50}{8} )^2 ≈150.6

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7 0
3 years ago
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yan [13]
<span>2 significant digits.
       
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3 0
3 years ago
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Mazyrski [523]
Hello ! I'm happy to help
the answer all revolves around this simple formula:
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