PROBLEM SOLVING Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each
survey. The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals. Find the
probability that at most 50 seals were observed during a randomly chosen survey.
The probability is

Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.

Step-by-step explanation:

We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each  survey.

The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.

Let X = numbers of seals observed

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

                             Z  =    ~ N(0,1)

where, = population mean numbers of seals = 73

            = standard deviation = 14.1

Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X 50 seals)

            P(X 50) = P( ) = P(Z -1.63) = 1 - P(Z < 1.63)    

                                                          = 1 - 0.94845 = 0.0516

The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.