Question #204751

f X is a normally distributed random variable with a mean of 45 and a standard deviation of 8, find the following probabilities: a. P(X >>> 50). b. P(X < 32). c. P(37 < X < 48). d. P(X = 45).


1
Expert's answer
2022-01-10T14:45:23-0500

μ=45σ=8\mu=45 \\ \sigma = 8

a.

P(X>50)=1P(X<50)=1P(Z<50458)=1P(Z<0.625)=10.7340=0.2660P(X>50) = 1 -P(X<50) \\ = 1 -P(Z< \frac{50-45}{8}) \\ = 1 -P(Z< 0.625) \\ = 1 -0.7340 \\ = 0.2660

b.

P(X<32)=P(Z<32458)=P(Z<1.625)=0.0520P(X<32) = P(Z< \frac{32-45}{8}) \\ = P(Z< -1.625) \\ = 0.0520

c.

P(37<X<48)=P(X<48)P(X<37)=P(Z<48458)P(Z<37458)=P(Z<0.375)P(Z<1)=0.64610.1586=0.4875P(37<X<48) = P(X<48) -P(X<37) \\ = P(Z< \frac{48-45}{8}) -P(Z< \frac{37-45}{8}) \\ = P(Z< 0.375) -P(Z< -1) \\ = 0.6461 -0.1586 \\ = 0.4875

d.

P(X=45) = 0

Because there is an uncountable infinite number of a value of X, therefore the probability of each individual value is zero.


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