Mathematics, 05.05.2020 04:02 lzrd3rik

Use the normal distribution to approximate the following binomial distribution: A convenience store owner claims that 55% of the people buying from her store, on a certain day of the week, buy coffee during their visit. A random sample of 35 customers is made. If the store owner's claim is correct, what is the probability that fewer than 23 customers in the sample buy coffee during their visit on that certain day of the week?

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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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

55% of the customers buy coffee, hence p = 0.55

Sample of 35 customers, hence n = 35

The mean and the standard deviation of the approximation are:

$\mu = np = 35(0.55) = 19.25$

$\sigma = \sqrt{np(1 - p)} = \sqrt{35(0.55)(0.45)} = 2.94$

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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{22.5 - 19.25}{2.94}$

Z = 1.1

Z = 1.1 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 link

Answer from: krizzle168

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 = $\sqrt{npq}$ = $\sqrt{8.6625}$ = 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( $\frac{X - np}{\sqrt{npq} } < \frac{20.5-np}{\sqrt{npq} }$ )

= P ( $Z < \frac{20.5-19.25}{2.9432}$ )

= P( Z < 0.42470) = 0.6628 (from standard normal table)

Answer from: Quest

$18.25 ($365 divided by 20)

Answer from: Quest

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation:

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Use the normal distribution to approximate the following binomial distribution: A convenience store...

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