Answer to Question #308874 in Statistics and Probability for Floryana

Question

1:The average amount of miles a car battery lasts for in a 2019 sedan is 37,452 miles with a standard deviation of

4,890 miles. If this distribution is normally distributed, answer the following questions:

A: What is the probability you select a battery it last longer than 40,000 miles?

B: What is the probability you select a battery, and it lasts between 32,000 miles and 44,000 miles?

C: Where does the top 18% of longest lasting batteries begin (at what mileage)?

Answer 1

Mean $\mu = 37,452$

Standard deviation $\sigma =4,890$

(a) Battery last longer than $40,000$

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

$Z=\dfrac{40,000-37,452}{4,890}=0.5211$

From normal distribution tables

P(Z of 0.5211) $=0.69847$

The battery lasting longer

$1-0.69846=0.30154$

(b) Battery lasting between $32,000$ and $44,000$

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

Z₁ $=\dfrac{32,000-37,452}{4,890}=-1.115$

Z₂ $=\dfrac{44,000-37,452}{4,890}=1.339$

From normal distribution tables

P₁ $=0.1335$

P₂ $= 1-0.90988 =0.09012$

$P(32,000\to44,000)=1-(0.1335+0.09012)$

$=0.77638$

(c) Where does top $18\%$ begin

Top $18\%$ begins at $100-18=82\%$

$P=0.82$

From normal distribution tables

Z(for P=0.82) $=0.92$

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

$X=σZ+\mu$

$X=0.92\times4,890+37,452$

$X=41,590$ miles