Assume that the random variable X has a normal distribution with a mean of 364 u

Question

Assume that the random variable X has a normal distribution with a mean of 364 units and a standard deviation of 8 units. Use this information to answer the remaining questions. What is the probability that the variable is less than 364? What is the probability that the variable is between 372 and 380? What is the probability that the variable is less than 352? What is the probability that the variable is more than 360? If the distribution of X is as described above what is the probability that X has values between 378 and 384.4? If the distribution of X is as described above, then what is the value of the 67^th percentile of the distribution? Find the value of X_o, such that P(X > X_o) = 0.0594. The price of a certain stock, X, is normally distributed with a mean of $50 and a standard deviation of $8. Use this information to answer the next four questions. Ninety-seven and one-half percent of the time this stock's price will be below what value? What is the probability that this stock's price will exceed $64? What is the probability that this stock's price will be between $44 and $60? For the variable, X, find the value of X_o, such that P(X > X_o) = 0.2946.

Explanation / Answer

Multiple questions. Few questions answered.

10).

Z value for 364, z=(364-364)/8 =0

P( x <364) = P(z < 0) =0.5

11).

Z value for 372, z=(372-364)/8 =1

Z value for 380, z=(380-364)/8 =2

P( 372<x<380) = P( 1<z<2)

=P( z <2) – P( z <1)

= 0.9772 - 0.8413

=0.1359

12).

Z value for 352, z=(352-364)/8 =-1.5

P( x <352) = P(z < -1.5) = 0.0668

13).

Z value for 360, z=(360-364)/8 =-0.5

P( x >360) = P(z > -0.5) = 0.6915

14).

Z value for 378, z=(378-364)/8 =1.75

Z value for 384.4, z=(384.4-364)/8 =2.55

P( 378<x<384.4) = P(1.75<z<2.55)

=P( z <2.55) – P( z <1.75)

= 0.9946 - 0.9599

=0.0347

15)

Z value for 67th percentile =0.44

X value = 364+0.44*8 = 367.52

16).

Z value for P(z >zo)=0.0594 is 1.56

X value = 364+1.56*8 = 376.48

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