Mathematics: 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

15.40Step-by-step explanation:...

A hotel claims that 85 % of its customers are very

KindaSmartPersonn

18.01.2022

(a) P (X = 7) = 0.387

(b) P (X > 7) = 0.2725

(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 $X\sim Bin(n=8, p=0.85)$ .

The probability distribution function for Binomial distribution is:

$P (X=x)={n\choose x}p^{x}(1-p)^{n-x}$

(a)

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

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

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:

$P (X7)=P(X=8)\\={8\choose 8}(0.85)^{8}(1-0.85)^{8-8}\\=1\times0.272491\times1\\=0.272491\\\approx0.2725$

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:

P(X

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:

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

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.

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