STAT Mutiple Choice Question Complete Solution

Question Details: #1288
STAT Mutiple Choice Question Complete Solution

  1. Find the following probabilities based on a standard normal variable Z. Use Table 1(Round your answers to 4 decimal places.)

 

 

 

 

  a.

P(Z > 0.74)

  

  b.

P(Z ≤ 1.92)

  

  c.

P(0 ≤ Z ≤ 1.62)

  

  d.

P(−0.90 ≤ Z ≤ 2.94)

  


 

  1. The cumulative probabilities for a continuous random variable X are P(X ≤ 10) = 0.42 and P(X ≤ 20) = 0.66. Calculate the following probabilities. (Round your answers to 2 decimal places.)

 

 

 

 

  a.

P(X > 10)

  

  b.

P(X > 20)

  

  c.

P(10 < X < 20)

  


 

  1. Let X be normally distributed with mean μ = 120 and standard deviation σ = 20. Use Table 1.

 

a.

Find P(X ≤ 86). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

 

  P(X ≤ 86)

  

 

b.

Find P(80 ≤ X ≤ 100). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

 

  P(80 ≤ X ≤ 100)

  

 

c.

Find x such that P(X ≤ x) = 0.40. (Round "z" value to 2 decimal places and final answer to nearest whole number.)

 

  x

  

 

d.

Find x such that P(X > x) = 0.90. (Round "z" value to 2 decimal places and final answer to 1 decimal place.)

 

  x

  


 

  1. Let X be normally distributed with mean μ = 2.5 and standard deviation σ = 2. Use Table 1.

 

a.

Find P(X > 7.6). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

 

  P(X > 7.6)

  

 

b.

Find P(7.4 ≤ X ≤ 10.6). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

 

  P(7.4 ≤ X ≤ 10.6)

  

 

c.

Find x such that P(X > x) = 0.025. (Round "z" value and final answer to 2 decimal places.)

 

  x

  

 

d.

Find x such that P(x ≤ X ≤ 2.5) = 0.4943. (Negative value should be indicated by a minus sign. Round "z" value and final answer to 2 decimal places.)

 

  x

  

 

  1. Find the following z values for the standard normal variable Z. Use Table 1. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

 

 

 

 

  a.

P(Z ≤ z) = 0.9744

  

  b.

P(Z > z) = 0.8389

  

  c.

P(−z ≤ Z ≤ z) = 0.95

  

  d.

P(0 ≤ Z ≤ z= 0.3315

  


 

 

 

 

Exhibit 6-1. You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F.

Refer to Exhibit 6-1. One sleeping bag you are considering advertises that it is good for temperatures down to 25°F. What is the probability that this bag will be warm enough on a randomly selected May night at the park?

0.8800

0.8106

0.1894

0.3106

It is known that the length of a certain product X is normally distributed with μ = 20 inches. How is the probability  Picture related to  Picture ?

\ is the same as \.

\ is smaller than \.

No comparison can be made with the given information.

\ is greater than \.

 

Let X be normally distributed with mean μ and standard deviation σ > 0. Which of the following is true about the z value corresponding to a given x value?

A positive z = (x - μ)/σ indicates how many standard deviations x is above μ.

All of the above.

A negative z = (x - μ)/σ indicates how many standard deviations x is below μ.

The z value corresponding to x = μ is zero.

 

Find the probability P(-1.96 ≤ Z ≤ 0).

0.0250

0.4750

0.5250

0.0500

 

The probability that a normal random variable is less than its mean is ___.

1.0

Cannot be determined

0.0

0.5

 

How many parameters are needed to fully describe any normal distribution?

2

1

3

4

 

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take between 2.5 and 3.5 hours to construct a soapbox derby car.

0.3085

0.6170

0.3830

0.6915

 

Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X ≤ x) = 0.9394.

126

1.55

374

-1.55

 

Exhibit 6-2. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of 3 ounces. Assume the amount of gold found per 1000 tons of dirt is normally distributed.

Refer to Exhibit 6-2. What is the probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated?

0.5918

0.0918

0.9082

0.4082

 

Top of Form

If X has a normal distribution with   and   , then the probability   can be expressed in terms of a standard normal variable Z as _______.

 

 

 

 

 

 

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Solution Details: #1266
STAT Mutiple Choice Question Complete Solution

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