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A popular retail store knows that the purchase amounts by its customers is a random variable that follows a normal distribution with a mean of $30 and a standard deviation of $9. what is the probability that a randomly selected customer will spend between $20 and $35 at this store? place your answer, rounded to 4 decimal places, in the blank. for example, 0.3456 would be a legitimate entry.

Answer :

carlosego
with continuity correction. 20-35 becomes 19.5 to 35.5 
 z1 = (19.5-30)/9 = -1.167 
 z2 = (35.5-30)/9 = 0.61 
 P(-1.167 < z < 0.61) = 0.6075
shinmin
We are going to use a continuity correction factor. When you use a normal distribution to get the estimated binomial distribution, you're going to have to use a continuity correction factor. So with this, the data will become from 20 - 35 it will now become 19.5 to 35.5.

So we are going to use the z table for this problem. So if we are looking for the z value, the formula would be: x – μ / σ

get the value of the first z value:

z1 = (19.5 - 30) / 9 = -1.167 

get the value of the second z value:

z2 = (35.5 - 30) / 9 = 0.61 

So getting the probability of both:

P(-1.167 < z < 0.61) =  P(Z ≤ b) – P(Z ≤ a) = 0.6075

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