Question

-. For what follows, Z denotes the standard normal random variable.

**(1) P(Z<1.26) is about**

(a) 0.7924

(b) 0.3962

(c) 0.1038

(d) 0.6038

(e) 0.8962

**(2) P(-0.54 < Z < 0.78) is about**

(a) 0.2054

(b) 0.0769

(c) 0.4877

(d) 0.2823

(e) 0.6877

-Assume the random variable X is normally distributed with mean 172 and standard deviation 10. Find the following probabilities.

**(3) P(X<160)**

(a) 0.8849

(b) 0.1151

(c) 0.9999

(d) 0.8078

(e) 0.8438

**(4) P(X>150)**

(a) 0.0139

(b) 0.9861

(c) 0.7910

(d) 0.8599

(e) 0.209

**(5) P(X>170 or X<140)**

(a) 0.4214

(b) 0.9993

(c) 0.58

(d) none of the above

Answer #1

1)

From Z score table

**P(Z<1.26) = 0.8962**

Option e is correct

2)

**P(-0.54 < Z < 0.78)** = P ( z < 0.78) -
P( -0.54 < z)

= 0.7823 - 0.2946

**= 0.4877**

**Option C is correct**

3)

**Mean = 172**

**S.D. = 10**

**P(X<160)**

From Z score table

*Z* = (X - ?) / ?

*Z* = (160 - 172) / 10

*Z* = -1.2

**P(X<160) = P( Z < -1.2) = 0.1151**

**option B**

**4)**

Z score at 150

*Z* = (X - ?) / ?

*Z* = (150 - 172) / 10

*Z* = -2.2

**P(X>150) = P ( Z > -2.2) = 0.9861?**

**option B**

**5)**

Z score at x=170

*Z* = (X - ?) / ?

*Z* = (170 - 172) / 10

*Z* = -0.2

Z score at x=140

*Z* = (X - ?) / ?

*Z* = (140 - 172) / 10

*Z* = -3.2

**P(X>170 or X<140) = P(X > -0.2 or X <
-3.2)**

**= P( X > -0.2) + P( X < -3.2 )**

**=** 0.5793? + 0.0007?

**= 0.58**

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