has an average daily expense of $75.00. the standard deviation is $15.00. the owner takes a sample of 64 bills. what is the probability the mean of his sample will be between $70.00 and $80.00? step 1. calculate a z-score for $70.00: step 2. give the probability for step 1: step 3. calculate the z-score for $80.00: step 4. give the probability for step 3: step 5. add the probabilities from steps 1&3:

Question

sillybilly123 · Answer

Trying to follow your STEPS  but they're a bit unhelpful, tbh

Population parameters are:

$\qquad  \mu = 75 \qquad \qquad \sigma = 15$

If a population is already (approximately) normally distributed (i.e. $X \sim N(\mu, \sigma)$), or if the sample size ($n$) of the sampling distribution is "sufficiently" large, then the **distribution of the sample means** $\bar x$ is normally distributed, i.e.:

$\qquad \bar X \sim N (\mu, \frac{\sigma}{\sqrt n} ) $

Because sample size $n = 64 \ge 30$ is looked at as large in most text books, you can rely on the latter (and so you are using the Central Limit Theorem).

ie the distribution of the sample mean has sample statistics related to population parameters, as follows:

$\qquad \mu_{\bar x} = \mu = 75 \qquad \qquad \sigma_{\bar x} = \frac{\sigma}{\sqrt n} = \frac{15}{\sqrt {64}} = 1.875$

Jay's **TINY** rectangular QC input window is "somewhat" unhelpful; but your next step is to transform the Sampling Distribution to a standardized Normal Distribution of the sample mean, using z-scores and the z-Table.

ie

$\bar z = \frac{\bar x - \mu}{\sigma_{\bar x}}$

And then use the fact that:

$\bar Z \sim N( 0, 1) $

sillybilly123 · Answer

$z(70) = \frac{70 - 75}{1.875} = - 2.6\dot 7 $
$z(80) = \frac{80 - 75}{1.875} = + 2.6\dot 7 $

sillybilly123 · Answer

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sillybilly123 · Answer

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slight correction

sillybilly123 · Answer

For the probabilities:

Looking at the LHS of that standard normal, you can use a typical spreadsheet formulation

->  NORMSDIST(-2.67) = 0.37%

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So the odds of being in the zone, using the symmetry, are:

$\mathbb P ( -2.6\dot 7  < \bar z <   2.6 \dot 7  ) = 1 - 2 \times 0.37\% = 99.26 \% $   [2 d.p. ]

But that's why your Step 2 is just wrong.  Published tables give the probability for positive z-scores.  You work from there to get negative scores.