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2 years ago

50 POINTS!!!!

Mathematics

1 answer:

Softa [21]2 years ago

3 0

Answer:

Step-by-step explanation:

Given that X - the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a standard deviation of 2.7 inches.

Height of male pilot = 74.2 inches

We have to find the percentile

X = 74.2

Corresponding Z score = 74.2-72.6 = 1.6

P(X<174.2) = P(Z<1.6) = 0.5-0.4452=0.0548=5.48%

i.e. only 5% are below him in height.

Thus the malepilot is in 5th percentile.

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AVprozaik [17]

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Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test t

bixtya [17]

Answer:

For this case we can find the critical value with the significance level $\alpha=0.05$ and if we find in the right tail of the z distribution we got:

$z_{\alpha}= 1.64$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of households with one person is significantly higher than 0.27

Step-by-step explanation:

We have the following dataset given:

$X= 43$ represent the households consisted of one person

$n= 125$ represent the sample size

$\hat p= \frac{43}{125}= 0.344$ estimated proportion of households consisted of one person

We want to test the following hypothesis:

Null hypothesis: $p \leq 0.27$

Alternative hypothesis: $p>0.27$

And for this case we can find the critical value with the significance level $\alpha=0.05$ and if we find in the right tail of the z distribution we got:

$z_{\alpha}= 1.64$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.344 -0.27}{\sqrt{\frac{0.27(1-0.27)}{125}}}=1.86$

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Move the -7 to the right, its sign will flip (do you know why?)
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