Question

1. Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of 29 months before certain electronic components deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 5 months, and the distribution of lifetimes is normal.

(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.)

%

(b) If Accrotime does not want to make refunds on more than 12% of
the watches it makes, how long should the guarantee period be (to
the nearest month)?

2.

Suppose *x* has a distribution with *μ* = 56 and
*σ* = 14.

(a) If random samples of size *n* = 16 are selected, can
we say anything about the *x* distribution of sample
means?

Yes, the *x* distribution is normal with mean
*μ*_{x} = 56 and
*σ*_{x} = 0.9.Yes, the *x*
distribution is normal with mean *μ*_{x} =
56 and *σ*_{x} =
14. Yes, the *x* distribution
is normal with mean *μ*_{x} = 56 and
*σ*_{x} = 3.5.No, the sample size is too
small.

(b) If the original *x* distribution is *normal*, can
we say anything about the *x* distribution of random samples
of size 16?

Yes, the *x* distribution is normal with mean
*μ*_{x} = 56 and
*σ*_{x} = 14.No, the sample size is too
small. Yes, the *x*
distribution is normal with mean *μ*_{x} =
56 and *σ*_{x} = 0.9.Yes, the *x*
distribution is normal with mean *μ*_{x} =
56 and *σ*_{x} = 3.5.

Find *P*(52 ≤ *x* ≤ 57). (Round your answer to four
decimal places.)

Answer #1

for normal distribution z score =(X-μ)/σx | |

mean μ= | 29 |

standard deviation σ= | 5 |

a)

percentage of total production will the company expect to replace :

probability =P(X<24)=(Z<(24-29)/5)=P(Z<-1)=0.1587 ~
15.87 % |

b)

for 12th percentile critical value of z=-1.17 | |

therefore
corresponding value=mean+z*std deviation=23
months |

2)a)

No, the sample size is too small.

b)

Yes, the *x* distribution is normal with mean
*μ*_{x} = 56 and
*σ*_{x} = 3.5

probability
=P(52<X<57)=P((52-56)/3.5)<Z<(57-56)/3.5)=P(-1.14<Z<0.29)=0.6141-0.1271=0.4870 |

(please try 0.4859 if this comes wrong)

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