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?

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:

In this problem:

The mean and the standard deviation of the approximation are:

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:

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

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 = =   = 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(  )

= P ()

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

step-by-step explanation:

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