Respuesta :
Answer:
The probability that fewer than 23 customers in the sample buy coffee during their visit on that certain day of the week is 0.6628
Step-by-step explanation:
In this question , we are to use the normal approximation to binomial distribution.
From the question, we identify the following
n= 35, p = 0.55 and q= 1-p = 1-0.55 = 0.45
npq = 35 ×0.55 × 0.45 = 8.6625
np = 35 × 0.55 = 19.25
nq = 35 × 0.45 = 15.75
The standard deviation σ = [tex]\sqrt{npq}[/tex] = [tex]\sqrt{8.6625}[/tex] = 2.9432
The normal distribution can be calculated using the formula below for the z-score(standard score)μσ
Z= (X - μ)/σ where μ is the mean
P(X < 21) = P(X< 21-0.5) = P(X<20.5) =P( [tex]\frac{X - np}{\sqrt{npq} } < \frac{20.5-np}{\sqrt{npq} }[/tex])
= P ([tex]Z < \frac{20.5-19.25}{2.9432}[/tex])
= P( Z < 0.42470) = 0.6628 (from standard normal table)
Using the normal approximation to the binomial, it is found that there is a 0.8643 = 86.43% probability that fewer than 23 customers in the sample buy coffee during their visit on that certain day of the week.
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
In this problem:
- 55% of the customers buy coffee, hence [tex]p = 0.55[/tex]
- Sample of 35 customers, hence [tex]n = 35[/tex]
The mean and the standard deviation of the approximation are:
[tex]\mu = np = 35(0.55) = 19.25[/tex]
[tex]\sigma = \sqrt{np(1 - p)} = \sqrt{35(0.55)(0.45)} = 2.94[/tex]
Using continuity correction, the probability that fewer than 23 customers in the sample buy coffee during their visit on that certain day of the week is P(X < 23 - 0.5) = P(X < 22.5), which is the p-value of Z when X = 22.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{22.5 - 19.25}{2.94}[/tex]
[tex]Z = 1.1[/tex]
[tex]Z = 1.1[/tex] has a p-value of 0.8643.
0.8643 = 86.43% probability that fewer than 23 customers in the sample buy coffee during their visit on that certain day of the week.
A similar problem is given at https://brainly.com/question/24261244
