Given that x is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability that x is between 47 and 54.

Let X be the normal variable with mean μ =50 and standard deviation σ =2

We have to find probability that x is between 47 and 54

P(47 < X < 54) = P(X < 54) - P(X < 47)

= -

= P(Z < 2) - P(Z < -1.5)

Using standard normal z score table to find probabilities we get

P(Z < 2) = 0.9772

P(Z < -1.5) = 0.0668

P(47 < X < 54) = P(Z < 2) - P(Z < -1.5)

= 0.9772 - 0.0668

P(47 < X < 54) = 0.9104

The probability that x is between 47 and 54 is 0.9104

its 4 because 2 times 2 is 4, if you double the number two you get four

< 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

⭐dhruv⭐