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MariettaO [177]
2 years ago
6

If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen

ts have heights (a) greater than 72 inches, (b) less than or equal to 64 inches, (c) between 65 and 71 inches inclusive, (d) equal to 68 inches? assume the measurements to be recorded to the nearest inch.
Mathematics
1 answer:
Lunna [17]2 years ago
3 0
Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

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Answer:

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

Step-by-step explanation:

Data given

Campus   Sample size     Mean    Population deviation

   1                 330               33                      8

   2                310                31                       7

\bar X_{1}=33 represent the mean for sample 1  

\bar X_{2}=31 represent the mean for sample 2  

\sigma_{1}=8 represent the population standard deviation for 1  

\sigma_{2}=7 represent the population standard deviation for 2  

n_{1}=330 sample size for the group 1  

n_{2}=310 sample size for the group 2  

\alpha Significance level provided  

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

P value  

Since is a one right tailed test the p value would be:  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

5 0
2 years ago
Given RT below, if S lies on RT such that the
defon

Answer:

(-2, -3)

Step-by-step explanation:

If a segment having extreme ends as (x_1,y_1) and (x_2,y_2) is divided by a point (x, y) in the ratio of m:n,

x = \frac{mx_2+nx_1}{m+n}

y = \frac{my_2+ny_1}{m+n}

Since, a line RT has extreme ends as R(-5, 3) and T(-1, -5) then a point S(x, y) which divides RT in the ratio of 3 : 1 will be,

x = \frac{3(-1)+1(-5)}{3+1}

  = \frac{-8}{4}

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y = \frac{3(-5)+1(3)}{3+1}

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Therefore, coordinates of the point S will be (-2, -3).

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2 years ago
Your friend claims that if a line has a slope less than 1, then any line perpendicular to it must have a positive slope. Is your
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Answer:

The friend's claim is not correct

Step-by-step explanation:

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal yo -1)

Example

A given line has a slope of m_1=1/2  ---> (is less than 1)

the slope of the line perpendicular to the given line is equal to

m_1*m_2=-1

substitute

(1/2)*m_2=-1

m_2=-2

so

the perpendicular line don't have a positive slope

therefore

The friend's claim is not correct

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Answer:

Heidi hass really kong hair and should cut it

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You find your dependent variable has a high degree of kurtosis. What is the first way to try to fix it?
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The first way to try to fix this is to apply logarithm to the observations on the dependent variable. This is going to make the dependent variable with high degree of kurtosis normal.

Note that sometimes, the resulting values of the variable will be negative. Do not worry about this, as it is not a problem. It does not affect the regression coefficients, it only affects the regression intercept, which after transformation, will be of no interest.

7 0
2 years ago
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