Question

gourmet chocolate sales: the sales of gourmet chocolates are normally distributed. using the cumulative normal distribution table, find the probability that sales on a given day are above the mean by more than 1.75 standard deviations. round the answer to at least four decimal places. the probability that sales on a given day are above the mean by more than 1.75 standard deviations is \boxed{}.

Explanation:

Step1: Recall Z - score concept

For a normal distribution, we want \( P(X > \mu+ 1.75\sigma)\). We can standardize this using the Z - score formula \( Z=\frac{X - \mu}{\sigma}\). Here, \( X=\mu + 1.75\sigma\), so \( Z=\frac{(\mu + 1.75\sigma)-\mu}{\sigma}=1.75\). We need to find \( P(Z > 1.75)\).

Step2: Use cumulative normal table

The cumulative normal distribution table gives \( P(Z\leq z)\). We know that \( P(Z > z)=1 - P(Z\leq z)\). From the cumulative normal table, \( P(Z\leq1.75) = 0.9599\) (we look up the value for \( z = 1.75\) in the standard normal table: for \( z = 1.7\) the cumulative probability is 0.9554, for \( z = 1.75\), we can calculate or look up the more precise value. The standard value for \( z = 1.75\) is 0.95994).

Step3: Calculate the probability

\( P(Z > 1.75)=1 - P(Z\leq1.75)=1 - 0.95994 = 0.04006\)

Answer:

\(0.0401\) (rounded to four decimal places)