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A hotel claims that 85 % of its customers are very satisfied with its service. Complete parts a through d below based on a random sample of eight customers.

a. What is the probability that exactly seven customers are very satisfied?

_______ (Round to four decimal places as needed.)

b. What is the probability that more than seven customers are very satisfied?

______ (Round to four decimal places as needed.)

c. What is the probability that less than six customers are very satisfied?

_______ (Round to four decimal places as needed.)

d. Suppose that of eight customers selected, two responded that they are very satisfied. What conclusions can be drawn about the hotel's claim?

The probability that 2 out of 8 customers are very satisfied is _______ assuming that 85 % of customers are very satisfied.
Based on this sample, it is (likely or not likely) ? that 85 % of customers are very satisfied. (Round to four decimal places as needed.)

Respuesta :

Answer:

(a) P (X = 7) = 0.387

(b) P (X > 7) = 0.2725

(c) P (X < 6) = 0.1052

(d) P (X = 2) = 0.0002

Based on this sample, it is not likely that 85 % of customers are very satisfied.

Step-by-step explanation:

Let X = number of customers very satisfied by the hotel's services.

The sample selected is of size, n = 8 and the probability that a customer is very satisfied is, p = 0.85.

The random variable [tex]X\sim Bin(n=8, p=0.85)[/tex].

The probability distribution function for Binomial distribution is:

[tex]P (X=x)={n\choose x}p^{x}(1-p)^{n-x}[/tex]

(a)

Compute the probability that exactly seven customers are very satisfied as follows:

[tex]P (X=7)={8\choose 7}(0.85)^{7}(1-0.85)^{8-7}\\=8\times 0.32058\times 0.85\\=0.384696\\\approx0.3847[/tex]

Thus, the probability that exactly seven customers are very satisfied is 0.3847.

(b)

Compute the probability that more than seven customers are very satisfied as follows:

[tex]P (X>7)=P(X=8)\\={8\choose 8}(0.85)^{8}(1-0.85)^{8-8}\\=1\times0.272491\times1\\=0.272491\\\approx0.2725[/tex]

Thus, the probability that more than seven customers are very satisfied is 0.2725.

(c)

Compute the probability that less than six customers are very satisfied as follows:

[tex]P(X<6)=1-P(X\geq 6) \\=1 - P(X=6)+P(X=7)+P(X=8)\\=1-{8\choose 6}(0.85)^{6}(1-0.85)^{8-6}-{8\choose 7}(0.85)^{7}(1-0.85)^{8-7}+{8\choose 8}(0.85)^{8}(1-0.85)^{8-8}\\=1-0.2376-0.3847-0.2725\\=0.1052[/tex]

Thus, the probability that less than six customers are very satisfied is 0.1052.

(d)

Compute the probability that of eight customers selected, two responded that they are very satisfied as follows:

[tex]P (X=2)={8\choose 2}(0.85)^{2}(1-0.85)^{8-2}\\=28\times0.7225\times0.000011\\=0.00023[/tex]

The probability that 2 out of 8 customers are very satisfied is 0.0002 assuming that 85 % of customers are very satisfied.

Based on this sample, it is not likely that 85 % of customers are very satisfied.

As per the question the hotel claims to have an 85% of the customers who are very satisfied with its services. The survey was done by a random sample of 8 customers.  Thus answer for the 7 client satisfaction is 0.387, more than seven is 0.2725, the P of less than six is 0.1052. The chance of 8 clients is 0.0002.

What is probability that exact customers are satisfied.

  1. The chances or the probability of the exact seven customers are staffed by the hotel staff and services is  P (X = 7) = 0.387.
  2. The chance for the more than seven customers are very satisfied by the hotel and staff is ) P (X > 7) = 0.2725.
  3. The chances for the probability of fewer than 6 customers are very satisfied is P (X < 6) = 0.1052.
  4. The chance of the 8 customers feels satisfied areas P (X = 2) = 0.0002. If 2 out of the 8 are assumed to be 85% very satisfied. Thus based on the above data not likely to all 85% of them are feeling satisfied.

Find out more information about the hotel's customer service.

brainly.com/question/8773716.

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