Answer:
0.2266
Step-by-step explanation:
Given that a survey indicates that shoppers spend an average of 26 minutes with a standard deviation of 8 minutes in your store and that these times are normally distributed.
i.e. if X is the time spent in the store then X is
N(26, 8)
The probability that a randomly selected shopper will spend less than 20 minutes in the store
=[tex]P(X<20)\\=P(Z<\frac{20-26}{8} \\=P(Z<-0.75)\\\\=0.2266[/tex]
Required prob = 0.2266