The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 1000 households from this region, let x be the number that do some or all of their banking on the Internet.

Mean = 310 Standard deviation = 14.63 z-score = -12.27

Find the approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130.

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, = 310/1000 or 0.31   and  Standard deviation, = 14.63/1000 = 0.01463 .

We know that Z = ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

 P(X <=0.13) = P(  <= ) = P(Z <= -12.303) = P(Z > 12.303)

Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.

So, after seeing the table we can say that this probability is approximately less than 0.0005% .

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step-by-step explanation: