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Oksana_A [137]
4 years ago
8

Rating Frequency Present Frequency Excellent 18 _______Very Good 25 _______Good 4 _______Fair 1 2Poor 2 4 Total 50 100What do th

ese summaries indicate about the overall customer satisfaction with the Delta flight? __________ % customers are satisfied with the Delta Flight Et either a good, very good, or excellent rating. Only ___________% of customers rated the Delta flight Fair or Poor. b. The online survey questionnaire enabled respondents to explain any aspect of the flight that Failed to meet expectations. Would this Se helpful information to a manager looking For ways to improve the overall customer satisfaction on Delta flights? Explain. Allowing survey respondents to explain their 5-point scale responses would provide helpful information tc managers looking for ways to improve customer satisfaction on Delta flights. The __________% respondents indicating that the flight failed to meet expectations would have the opportunity to provide detailed information about their expectations.
Mathematics
1 answer:
nata0808 [166]4 years ago
6 0

Answer:

a) Rating Frequency Percent Frequency

Excellent 18 ___36%____Very Good 25 ___50%____Good 4 ___8%____Fair 1   2% Poor 2  4%  Total 50 100 What do these summaries indicate about the overall customer satisfaction with the Delta flight? ____94%______ % customers are satisfied with the Delta Flight Et either a good, very good, or excellent rating. Only ____6%_______% of customers rated the Delta flight Fair or Poor.

b. The online survey questionnaire enabled respondents to explain any aspect of the flight that Failed to meet expectations. Would this be helpful information to a manager looking for ways to improve the overall customer satisfaction on Delta flights? Explain.

Allowing survey respondents to explain their 5-point scale responses would provide helpful information to managers looking for ways to improve customer satisfaction on Delta flights. The ____6______% respondents indicating that the flight failed to meet expectations would have the opportunity to provide detailed information about their expectations.

Step-by-step explanation:

a) Data and Calculations:

                  Rating Frequency     Percent Frequency

Excellent                18                       ____36%___

Very Good            25                       ____50%___

Good                       4                       ____  8%___

Fair                          1                        ____  2%___

Poor                        2                       _____4% __

Total                     50                               100%

b) Percentage Frequencies are obtained as follows:

Excellent = 18/50 * 100 = 36%

Very Good = 25/50 * 100 = 50%

Good = 4/50 * 100 = 8%

Fair = 1/50 * 100 = 2%

Poor = 2/50 * 100 = 4%

c) Good, very good, and excellent = 47/50 * 100 = 94%

d) Fair or Poor rating = 3/50 * 100 = 6%

 

e) Moreover, dissatisfied customers can be sources of great information on the performance of a good or service.  Though, with only 6% of customers giving a dissatisfaction rating with the Delta flights, it is still important to dive deep to unearth the reasons for their dissatisfaction as this may provide valuable information to improve the services.  Usually, this class of customers is the most sincere in most cases, unless their dissatisfaction can be proved to be attitudinal on their own part.

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Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal
Usimov [2.4K]

Answer:

a) We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

b) We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

Step-by-step explanation:

Part a

We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

Part b

We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

7 0
4 years ago
Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.
MrMuchimi
A random variable following a binomial distribution over n trials with success probability p has PMF

f_X(x)=\dbinom nxp^x(1-p)^{n-x}

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1

The mean is given by the expected value of the distribution,

\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}
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\mathbb E(x)=np=126\times0.27=34.02

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The standard deviation is just the square root of the variance, which is

\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834
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3 years ago
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HACTEHA [7]

Answer:

V =41.41³

A = 94.41²

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V = 64³

SA =113.32²

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SA = 378.57²

Step-by-step explanation:

Picture 2 = a = 1/2 base = 3.5 x 3.5 = 12.25  b= 5 x 5 = 25

c²= a² + b² = 3.5² + 5²

c ²= √12.25 + √25

c ²= √ 37.5 = 6.12372435696

c ² = 6.1237   missing side

Picture 1 + 2  formula SA = bh + (s1 + s2 + s3)H

V =  V= 1/2 b x h  h x SA

Picture 3 + 4  formula SA= a²+ 2a   a² / 4 + h²    V= a²  h/3

5 0
3 years ago
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Sever21 [200]

Answer:

no lol but what do u need help with

8 0
4 years ago
I don’t know how to do this. can someone please help me :)
Kaylis [27]
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The 3rd selection is appropriate.
8 0
3 years ago
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