Given that z is a standard normal random variable, find z for each situation (to 2 decimals). The area to the left of z is 0.6591. The area to the right of z is 0.3409.

Five percent of U.S. employees who are late for work blame oversleeping. You randomly select four U.S. employees who are late for work and ask them whether they blame oversleeping. The random variable represents the number of U.S. employees who are late for work and blame oversleeping. Find the mean of the binomial distribution. μ= (Round to the nearest hundredth as needed.)

Use the standard normal table to find the area to the left of z=−0.34. The area to the left of z=−0.34 is

Let X₁, X2,..., be a sequence of independent and identically distributed random variables, each with a mean value x and variance ok, both of which are assumed to be finite. Let N be a discrete random variable independent of X₁, X2,..., and assuming values in the set {0, 1, 2, ...}, with mean value N and variance ok (both are finite). Form the random-compound sum Y=1 Xk, with the convention Y = 0 whenever N = 0.