What is the z-score of sample X, if $n=100,\mu =61,St.Dev.=43,\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}E\left[X\right]=98$?

proximumha
2022-08-12
Answered

What is the z-score of sample X, if $n=100,\mu =61,St.Dev.=43,\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}E\left[X\right]=98$?

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Emely English

Answered 2022-08-13
Author has **16** answers

The z-score for a sample mean is calculated by the formula

$z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$\overline{x}=E\left(x\right)=98$

$\mu =61$

$\sigma =43$

$n=100$

Therefore, $z=\frac{98-61}{\frac{43}{\sqrt{100}}}$

$=\frac{37}{4.3}=8.6$

$z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$\overline{x}=E\left(x\right)=98$

$\mu =61$

$\sigma =43$

$n=100$

Therefore, $z=\frac{98-61}{\frac{43}{\sqrt{100}}}$

$=\frac{37}{4.3}=8.6$

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asked 2022-08-19

A die is rolled 60 times.

Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times.

The exact chance that the face with six spots appears 9, 10, or 11 times?

my solution; in one roll of the die $P(6\text{spots})=\frac{1}{6}$ In 60 rolls of the die the expected number of times the face with 6 spots appears is given by $E(X)=\text{Mean}=\frac{60}{6}=10$ times.

$SE(6spots)=SD=SQR(\frac{60}{6}\cdot \frac{5}{6})=2.8868$

Applying the normal approximation to the chance that the face with six spots appears 10 times, I obtain the following: mean = 10

$\text{Standard deviation}=2.8868$

Using the continuity correction of 0.5 the following z-scores are found: ${z}_{1}=(9.5-10)/2.8868=-0.1732$ and ${z}_{2}=(10.5-10)/2.8868=0.1732$ Subtracting the cumulative probabilities for the two z-scores gives $0.5688-0.4312=0.1376$ which is the approximate chance that the face with six spots appears 10 times. find the six spots appears between 9,10, or 11?

The exact chance that the face with six spots appears 10 times is found from

$60C10{\left(\frac{1}{6}\right)}^{10\times}{\left(\frac{5}{6}\right)}^{50}=0.137$ find six spots appears 9, 10 or 11 times?

Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times.

The exact chance that the face with six spots appears 9, 10, or 11 times?

my solution; in one roll of the die $P(6\text{spots})=\frac{1}{6}$ In 60 rolls of the die the expected number of times the face with 6 spots appears is given by $E(X)=\text{Mean}=\frac{60}{6}=10$ times.

$SE(6spots)=SD=SQR(\frac{60}{6}\cdot \frac{5}{6})=2.8868$

Applying the normal approximation to the chance that the face with six spots appears 10 times, I obtain the following: mean = 10

$\text{Standard deviation}=2.8868$

Using the continuity correction of 0.5 the following z-scores are found: ${z}_{1}=(9.5-10)/2.8868=-0.1732$ and ${z}_{2}=(10.5-10)/2.8868=0.1732$ Subtracting the cumulative probabilities for the two z-scores gives $0.5688-0.4312=0.1376$ which is the approximate chance that the face with six spots appears 10 times. find the six spots appears between 9,10, or 11?

The exact chance that the face with six spots appears 10 times is found from

$60C10{\left(\frac{1}{6}\right)}^{10\times}{\left(\frac{5}{6}\right)}^{50}=0.137$ find six spots appears 9, 10 or 11 times?

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