Question

The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will last between 2.8 and seven years. (a) Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot (b) Give the probability statement and the probability. (Enter exact numbers as integers, fractions, or decimals for the probability statement. Round the probability to four decimal places.) P < x < =

Answer 1

Using the normal distribution, we have that:

a) The sketch of the situation is given at the end of this answer.

b) The probability is:

$P(2.8 \leq X \leq 7) = 0.8284$

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 4.1 years, thus $\mu = 4.1$.
• Standard deviation of 1.3 years, thus $\sigma = 1.3$.

Item a:

The part between 2.8 and 7 years is shaded on the sketch given at the end of this answer.

Item b:

The probability is the p-value of Z when X = 7 subtracted by the p-value of Z when X = 2.8, thus:

X = 7:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7 - 4.1}{1.3}$

$Z = 2.23$

$Z = 2.23$ has a p-value of 0.9871.

X = 2.8:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2.8 - 4.1}{1.3}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.9871 - 0.1587 = 0.8284, thus:

$P(2.8 \leq X \leq 7) = 0.8284$

A similar problem is given at brainly.com/question/25151638