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frutty [35]
3 years ago
12

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 100

0 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.
Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

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, \mu = 310/1000 or 0.31   and  Standard deviation, \sigma = 14.63/1000 = 0.01463 .

We know that Z = \frac{X-\mu}{\sigma} ~ 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( \frac{X-\mu}{\sigma}  <= \frac{0.13-0.31}{0.01463} ) = 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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