Question

A bagel shop wants to estimate their daily demand for their popular poppy seed bagel. The shop owner determined that the daily demand follows the normal probability model with mean 300 bagels and standard deviation 60. The probability that the number of poppy seed bagels sold on a particular day exceeds 400 is CLOSEST to:

Answer 1

Answer:

Probability that the number of poppy seed bagels sold on a particular day exceeds 400 is 0.0475.

Step-by-step explanation:

We are given that the shop owner determined that the daily demand follows the normal probability model with mean 300 bagels and standard deviation 60.

Let X = daily demand of poppy seed bagels

The z-score probability distribution for normal distribution is given by;

           Z = $\frac{ X-\mu}{\sigma}} }$  ~ N(0,1)

where, $\mu$ = mean bagels = 300

            $\sigma$ = standard deviation = 60

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the number of poppy seed bagels sold on a particular day exceeds 400 is given by = P(X > 400)

 P(X > 400) = P( $\frac{ X-\mu}{\sigma}} }$ > $\frac{ 400-300}{60}} }$ ) = P(Z > 1.67) = 1 - P(Z $\leq$ 1.67)

                                                   = 1 - 0.95254 = 0.0475                       

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.67 in the z table which has an area of 0.95254.

Hence, the probability that the number of poppy seed bagels sold on a particular day exceeds 400 is 0.0475.