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horsena [70]
2 years ago
5

if you invest 1500 today in a bank that gives you 5% annual interest eats, which of these items can you buy in two years

Mathematics
1 answer:
Minchanka [31]2 years ago
5 0
1500 dived by 5% should give u ur answer
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if 1343243253254353462534543254324534234542345134 times 23423543254353454325234543543543254365645654654 minus 454235435435425243
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3.14635164e91

Step-by-step explanation:

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In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hir
Ede4ka [16]

Answer:

Explained below.

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:

 \mu_{\hat p}= p

The standard deviation of this sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

(a)

The sample selected is of size <em>n</em> = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0204^{2}).

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.96

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0306^{2}).

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.31

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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3 years ago
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