Question 1:

A manufacturer knows that their items have a normally distributed length, with a mean of 18 inches, and standard deviation of 4.9 inches.

If one item is chosen at random,

  1. what is the probability that it is less than 20.5 inches long?

  2. what is the probability that it is greater than or equal to 21.7 inches long?

  3. what is the probability that it is between 20.5 and 21.7?

(Round your z-value to 2 decimal places and your final answer to 4 decimal places)

Solution 1:

  1. \(z=\frac{x-\mu}{\sigma}=\) 0.51. \(P(x<\) 20.5) \(=\) \(P(z<\) 0.51) \(=\) 0.695.

  2. \(z=\frac{x-\mu}{\sigma}=\) 0.76. \(P(x>\) 21.7) \(=\) \(P(z>\) 0.76) \(=1-P(z \leq\) 0.76) \(=\) 1- 0.7764 \(=\) 0.2236.

  3. \(z=\frac{x-\mu}{\sigma}=\) 0.51. \(P(x<\) 20.5) \(=\) \(P(z<\) 0.51) \(=\) 0.695.

    \(z=\frac{x-\mu}{\sigma}=\) 0.24. \(P(x<\) 21.7) \(=\) \(P(z<\) 0.24) \(=\) 0.5948.

    \(P(z<\) 0.24) \(-\) \(P(z<\) 0.24) \(=\) 0.5948 \(-\) 0.5948 \(=\) -0.1002.

Question 2:

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.8 years, and standard deviation of 2.6 years.

If you randomly purchase one item,

  1. what is the probability that it will last less than 22.3 years?

  2. what is the probability that it will last at least (i.e. greater than or equal) 13.5 years?

  3. what is the probability that it will last between 13.5 and 22.3 years?

(Round your z-value to 2 decimal places and your final answer to 4 decimal places)

Solution 2:

  1. \(z=\frac{x-\mu}{\sigma}=\) 3.65. \(P(x<\) 22.3) \(=\) \(P(z<\) 3.65) \(=\) 0.9999.

  2. \(z=\frac{x-\mu}{\sigma}=\) 0.27. \(P(x>\) 13.5) \(=\) \(P(z>\) 0.27) \(=1-P(z \leq\) 0.27) \(=\) 1- 0.6064 \(=\) 0.3936.

  3. \(z=\frac{x-\mu}{\sigma}=\) 0.27. \(P(x<\) 13.5) \(=\) \(P(z<\) 0.27) \(=\) 0.6064.

    \(z=\frac{x-\mu}{\sigma}=\) 3.38. \(P(x<\) 22.3) \(=\) \(P(z<\) 3.38) \(=\) 0.9996.

    \(P(z<\) 3.38) \(-\) \(P(z<\) 3.38) \(=\) 0.9996 \(-\) 0.9996 \(=\) 0.3932.