According to Net Market Share, Microsoft's Internet Explorer browser has 53.4% of the global market. A random sample of 70 users was selected. What is the probability that 32 or more from this sample used Internet Explorer as their browser?

Probability that 32 or more from this sample used Internet Explorer as their browser is 0.9015.

Step-by-step explanation:

We are given that according to Net Market Share, Microsoft's Internet Explorer browser has 53.4% of the global market.

A random sample of 70 users was selected.

Let = sample proportion of users who used Internet Explorer as their browser.

The z score probability distribution for sample proportion is given by;

                            Z  =    ~ N(0,1)

where, p = population proportion of users who use internet explorer = 53.4%

            = sample proportion = = 0.457

           n = sample of users = 70

Now, probability that 32 or more from this sample used Internet Explorer as their browser is given by = P( 0.457)

      P( 0.457) = P( ) = P(Z -1.29)

                            = P(Z 1.29) = 0.9015

The above probability is calculated by looking at the value of x = 1.29 in the z table which has an area of 0.9015.

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we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

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