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When using the CLT, it is very easy to end up with rounding errors depending on how you enter the standard deviation of the sampling distribution. Remember that the formula is:
[tex]\[
\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}.
\][/tex]

Suppose you take a random sample of 7 values from a normally distributed population with mean 19.9 and standard deviation 6.2.

a. Find [tex]\(\sigma_{\bar{x}}\)[/tex] and round to one decimal: [tex]\(\square\)[/tex]

Now use the normalcdf (TI) or NormalDist.Cdf (ClassCalc) command to find the following, rounding each answer to 4 decimals.

b. Find [tex]\(P(\bar{x}\ \textless \ 19)\)[/tex] using your rounded answer to part a: [tex]\(\square\)[/tex]

c. Find [tex]\(P(\bar{x}\ \textless \ 19)\)[/tex] by entering [tex]\(\frac{6.2}{\sqrt{7}}\)[/tex] instead: [tex]\(\square\)[/tex]

Even though the answers in parts b and c are close, on later problems the system might grade your answer as incorrect if you use a rounded value for [tex]\(\sigma_{\bar{x}}\)[/tex] like you did in part b. So be sure to enter an "exact" value, as you did in part c, to avoid rounding errors.

You should follow the same process when using the invNorm (TI) or NormalDist.inverseCdf (ClassCalc) command.

d. Find the value of [tex]\(a\)[/tex] (to 2 decimals) such that [tex]\(P(\bar{x}\ \textless \ a)=0.62\)[/tex]. [tex]\(\square\)[/tex]

Sagot :

GinnyAnswer
Let's solve the problem step-by-step based on the given details:

### Part a
To find [tex]\(\sigma_{\bar{x}}\)[/tex], we use the formula:

[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]

Given:
- [tex]\(\sigma = 6.2\)[/tex]
- [tex]\(n = 7\)[/tex]

First, we calculate the denominator:

[tex]\[ \sqrt{7} = 2.6457513110645907 \][/tex]

Then, we divide the population standard deviation by this value:

[tex]\[ \sigma_{\bar{x}} = \frac{6.2}{2.6457513110645907} \approx 2.343 \][/tex]

Rounding to one decimal place:

[tex]\[ \sigma_{\bar{x}} \approx 2.3 \][/tex]

### Part b
We need to find [tex]\(P(\bar{x} < 19)\)[/tex] using the rounded [tex]\(\sigma_{\bar{x}} = 2.3\)[/tex].

First, we calculate the z-score:

[tex]\[ z = \frac{19 - 19.9}{2.3} \approx -0.3913 \][/tex]

Using normal distribution tables or a cumulative distribution function (CDF) calculator, we find the probability corresponding to this z-score:

[tex]\[ P(\bar{x} < 19) \approx 0.3478 \][/tex]

### Part c
Now, we find [tex]\(P(\bar{x} < 19)\)[/tex] using the exact value of [tex]\(\sigma_{\bar{x}}\)[/tex]:

[tex]\[ \sigma_{\bar{x}} = \frac{6.2}{\sqrt{7}} \approx 2.343 \][/tex]

Calculate the z-score again:

[tex]\[ z = \frac{19 - 19.9}{2.343} \approx -0.3847 \][/tex]

Using CDF calculator for the exact z-score, we get:

[tex]\[ P(\bar{x} < 19) \approx 0.3505 \][/tex]

### Part d
We need to find the value of [tex]\(a\)[/tex] such that [tex]\(P(\bar{x} < a) = 0.62\)[/tex].

First, find the z-score for which the cumulative probability is 0.62. Using the Z-table or inverse CDF:

[tex]\[ z \approx 0.305 \][/tex]

Using the exact [tex]\(\sigma_{\bar{x}}\)[/tex], we calculate [tex]\(a\)[/tex]:

[tex]\[ a = 19.9 + 0.305 \times 2.343 \approx 19.9 + 0.714315 \approx 20.62 \][/tex]

Hence, gathering all the results:

- a. [tex]\( \sigma_{\bar{x}} \approx 2.3 \)[/tex]
- b. [tex]\( P(\bar{x} < 19) \approx 0.3478 \)[/tex]
- c. [tex]\( P(\bar{x} < 19) \approx 0.3505 \)[/tex]
- d. [tex]\( a \approx 20.62 \)[/tex]
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