Question

Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of 26 months before certain electronic components deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 4 months, and the distribution of lifetimes is normal. A button hyperlink to the SALT program that reads: Use SALT. (a) If Accrotime guarantees a full refund on any defective watch for 2 years after purchase, what percentage of total production will the company expect to replace? (Round your answer to two decimal places.) Incorrect: Your answer is incorrect. % (b) If Accrotime does not want to make refunds on more than 6% of the watches it makes, how long should the guarantee period be (to the nearest month)? months

Answer 1

Answer:

a

   $P(X < 24 )= 21.186\%$  

b

    $x = 19.78 \ months$

Step-by-step explanation:

From the question we are told that

 The mean is   $\mu = 26 \ months$

 The standard deviation is  $\sigma = 4 \ months$

 Generally 2 year is  equal to  24 months

Generally the percentage of total production will the company expect to replace is mathematically represented as

      $P(X < 24 )= P(\frac{X - \mu }{ \sigma} < \frac{24 - 26}{4} )$

Generally  $\frac{X - \mu}{\sigma } =Z (The \ standardized \ value \ of \ X )$

     $P(X < 24 )= P(Z < -0.8 )$

Generally from the z-table  

       $P(Z < -0.8) = 0.21186$

So

       $P(X < 24 )= 0.21186$

Converting to percentage

      $P(X < 24 )= 0.21186 * 100$

=>    $P(X < 24 )= 21.186\%$  

Generally the duration that should be the guarantee period if  Accrotime does not want to make refunds on more than 6% is mathematically evaluated as

    $P(X < x) = P(\frac{X - \mu }{\sigma} < \frac{x - 26}{4} )= 0.06$

=> $P(X < x) = P(Z < \frac{x - 26}{4} )= 0.06$

From the normal distribution table the z-score for  0.06 at the lower tail  is

       $z = -1.555$

So

    $\frac{x - 26}{4} = -1.555$

=> $x = 19.78 \ months$