Question

use the appropriate normal distribution to approximate the resulting binomial distributions. 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 22 customers in the sample buy coffee during their visit on that certain day of the week? a) 0.7764 b) 0.7967 c) 0.7224 d) 0.8413 e) 0.8238 f) none of the above.

Answer 1

The probability that fewer than 22 customers in the sample buy coffee during their visit is 0.7764.

What is a normal approximation to the binomial?

When we utilize a continuous distribution (the normal distribution) to approximate a discrete distribution, we use the normal approximation to the binomial (the binomial distribution). The Central Limit Theorem states that if the sample size is high enough, the sampling distribution of the sample means approximates normality.

In a normal distribution with mean and standard deviation, the z-score of a measure X is given by:

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

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

Sample of 35 customers, n = 35

mean  $\mu$ = np = 35*0.55 = 19.25

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

So,  the probability that fewer than 23 customers in the sample buy coffee is

P(X < 22 - 0.5) = P(X < 21.5)

= $P(\frac{X-\mu}{\sigma} < \frac{21.5-19.25}{2.94})$

= P(Z<0.765) = 0.7764

Hence, the probability that fewer than 22 customers in the sample buy coffee during their visit is 0.7764.

To learn more about a normal approximation to the binomial

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