116k views
5 votes
1 pts

Question 5
The size of gasoline tanks in cars is normally distributed with a mean size of 24.8 gallons and a standard
deviation of 6.2 gallons. What percent of tanks are less than 31 gallons. Round answer to the nearest
percent
84%
71%
16%
20%

User Sabir Khan
by
8.5k points

1 Answer

0 votes

Answer:


P(X<31)=P((X-\mu)/(\sigma)<(31-\mu)/(\sigma))=P(Z<(31-24.8)/(6.2))=P(z<1)

And we can find this probability using the normal standard distribution or excel and we got:


P(z<1)= 0.84

And if we convert this into % we got 84% so then the best solution would be:

84%

Explanation:

Let X the random variable that represent the size of gasoline tanks of a population, and for this case we know the distribution for X is given by:


X \sim N(24.8,6.2)

Where
\mu=24.8 and
\sigma=6.2

We are interested on this probability


P(X<31)

And we can use the z score formula given by:


z=(x-\mu)/(\sigma)

Using the last formula we got:


P(X<31)=P((X-\mu)/(\sigma)<(31-\mu)/(\sigma))=P(Z<(31-24.8)/(6.2))=P(z<1)

And we can find this probability using the normal standard distribution or excel and we got:


P(z<1)= 0.84

And if we convert this into % we got 84% so then the best solution would be:

84%

User Yanto
by
8.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories