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nika2105 [10]
2 years ago
6

The distribution of heights of 6-year-old girls is approximately normally distributed with a mean of 46.0 inches and a standard

deviation of 2.7 inches. Aaliyah is 6 years old, and her height is 0.96 standard deviation above the mean. Her friend Jayne is also 6 years old and is at the 93rd percentile of the height distribution. At what percentile is Aaliyah’s height, and how does her height compare to Jayne’s height?

Mathematics
1 answer:
Gelneren [198K]2 years ago
7 0

Answer:

Aaliyah's height is the 84th percentile and the relationship that there exists between Aaliyah's height and between Jayne's height is that Aaliyah's height is less than Jayne's height.

Step-by-step explanation:

Below we can observe the empirical rule with a mean of 46.0 inches and a standard deviation of 2.7 inches. We have that 48.7 inches represents one standard deviation above the mean, so, we can consider that Aaliyah's height is the 84th percentile. On the other side, Jayne's height is the 93rd percentile of the height distribution. Therefore, the relationship that there exists between Aaliyah's height and between Jayne's height is that Aaliyah's height is less than Jayne's height.

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4.05 – 4 = 0.05

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4.021 – 4 = 0.021

 

We can see that the value of 3.98g has the lowest difference and is therefore the closest to 4g. Hence the most precise is:

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2 years ago
Test the given claim. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, and then state the concl
never [62]

Answer:

a) Failed to reject the null hypothesis (P-value=0.09).

b) The 95% CI for the difference in proportions is:

-0.0599\leq\pi_1-\pi_2\leq0.0124

Step-by-step explanation:

a) We have to perform a hypothesis test for the difference of proportions.

The null and alternative hypothesis are:

H_0: \pi_1\geq\pi_2\\\\H_1: \pi_1

The significance level is 0.05.

The proportion of the passenger cars owners is:

p_1=\frac{239}{2142} =0.1116

The proportion of commercial truck owners is:

p_2=\frac{54}{399}=0.1353

The weigthed average p is

p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{239+54}{2142+399}=0.1153

The estimated standard deviation is

s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2}} =\sqrt{\frac{0.1153(1-0.1153)}{2142}+\frac{0.1153(1-0.1153)}{399}} =0.0174

We can calculate the z-value as:

z=\frac{\Delta p}{s}=\frac{0.1116-0.1353}{0.0174}=-1.362

The P-value for z=-1.362 is P=0.0866.

The P-value (0.09) is greater than the significance level (0.05), so it failed to reject the null hypothesis. There is no enough evidence to prove that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars.

b) We can construct a 95% CI, according to the significance level of 0.05.

The z-value for this CI is 1.96.

We have to recalculate the standard deviation:

\sigma=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2}} =\sqrt{\frac{0.1116(1-0.1116)}{2142} +\frac{0.1353(1-0.1353)}{399}} =0.0184

The lower limit is then:

LL=(p_1-p_2)-z*\sigma=(0.1116-0.1353)-1.96*0.0184=-0.0238-0.0361\\\\LL=-0.0599

The upper limit is:

UL=(p_1-p_2)+z*\sigma=(0.1116-0.1353)+1.96*0.0184=-0.0238+0.0361\\\\UL=0.0124

The 95% CI for the difference in proportions is:

-0.0599\leq\pi_1-\pi_2\leq0.0124

In this case, we can conclude that the difference between the proportions, with 95% confidence, can still be equal or greater than zero, meaning that it is possible passenger car owners violate laws more than truck owners.

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Now, you can use this song to help remember how to do these problems.

Cross off the sides till you get to the center. 1 is good, 2 is bad. (If you get two numbers in the middle) add then divide by two.

Cross off 21, then 89, then 33, then 79, then 33, then 67. Now, you're left with 42 in the middle. That is your median. Hope this helps! :)
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Answer:

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Answer:

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Step-by-step explanation:

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