Americans spend an average of 2 hours per day online. If the standard deviation is 23 minutes, find the range in which at least 75% of the data will lie.

75% of the data lies on the interval [1.5592, 2.4408]

Step-by-step explanation:

Let X be the amount of hours a random american spends online per day. X has normal distribution with parameters μ = 2 and σ = 23/60 = 0.383 . The standarization of X, W is

Which has distribution N(0,1). The values of the cummulative distribution function of W are tabulated and they can be found in the attached file.

We will find and interval centered in the mean 2 in which the probability that a random american belongns to that interval is 0.75. The interval has the form

I = [2-Z , 2+Z]

As a consecuence

Since W is symmetric respect to 0, then

Thus, the last expression is equal to . We conclude that

Looking at the table, we will find that Z/0.383 = 1.15, then Z = 1.15*0.383 = 0.4408. Therefore, the interval we are looking for is [2-0.4408, 2+0.4408] = [1.5592, 2.4408].

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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