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2 years ago

The mean and standard deviation of a normal distribution are 150 and 12

Mathematics

1 answer:

fomenos2 years ago

6 0

a) (150 - 138) / 12 = 1 deviation or 34%

(162 - 150) / 12 = 1 deviation or 34%

34% + 34% = 68%

b) (150 - 126) / 12 = 2 deviations or 47.5%

(174 - 150) / 12 = 2 deviations or 47.5%

47.5% + 47.5% =95%

c) (150 - 126) / 12 = 2 deviations or 47.5%

(162 - 150) / 12 = 1 deviation or 34%

47.5% + 34% = 81.5%

d) (162 - 150) / 12 = 1 deviation or 34%

(174 - 150) / 12 = 2 deviations or 47.5%

This problem is a little different. We have to subtract 34% from 47.5%

47.5% - 34% = 13.5%

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Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean iron concentration = 0.802 cc/cubic meter

s = sample standard deviation = 0.093

n = number of water samples = 27

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Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.706 < $t_2_6$ < 1.706) = 0.90 {As the critical value of t at 26 degree of

freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.706) = 0.90

P( $-1.706 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.706 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ , $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ]

= [ $0.802 -1.706 \times {\frac{0.093}{\sqrt{27} }$ , $0.802 +1.706 \times {\frac{0.093}{\sqrt{27} }$ ]

= [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

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