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____ [38]
3 years ago
10

Decrease £19064.67 by 19.5%

Mathematics
2 answers:
8_murik_8 [283]3 years ago
8 0
Answer = £15347.06

100 - 19.5 = 80.5

80.5 / 100 = 0.805

19064.67 x 0.805 = £15347.06
nekit [7.7K]3 years ago
8 0

Answer:15347.07

Step-by-step explanation:

Firstly we have to find 19.5%of 19064.67

Which is 19.5/100×19064.67/1=(19.5×19064.67)÷(100×1)

=371761.1÷100

=3717.6

Decreasing 19.5%of 19064.67 will now be

=19064.67-3717.6

=15347.07

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a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=400 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=80 represent the population standard deviation

n=15 represent the sample size  

2) Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58

Now we have everything in order to replace into formula (1):

400-2.58\frac{80}{\sqrt{15}}=346.708    

400+2.58\frac{80}{\sqrt{15}}=453.292

So on this case the 99% confidence interval would be given by (346.708;453.292)    

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that t_{\alpha/2}=2.98

Now we have everything in order to replace into formula (1):

400-2.98\frac{80}{\sqrt{15}}=338.445    

400+2.98\frac{80}{\sqrt{15}}=461.555

So on this case the 99% confidence interval would be given by (338.445;461.555)    

4 0
3 years ago
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