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The time is takes a police department to arrive at the scene of a crime is normally distributed with a mean of 4.5 minutes and a standard deviation of 0.8 minutes. What percent of time will the police take 4 minutes or less to arrive? Calculate the z-score, and use the table on page 447 of the textbook

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Given:


Mean\text{ }\mu\text{ =4.5, standard deviation }\sigma\text{ =0.8, and x =4.}

Required:

We need to find the percentage of the time will the police take 4 minutes or less to arrive.

Step-by-step explanation:

Consider the z-score formula.


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


Substitute\text{ }\mu\text{ =4.5, }\sigma\text{ =0.8, and x =4 in the formula.}
z=(4-4.5)/(0.8)
z=-0.625

P-value from Z-Table


P(x<4)=0.26599

Multiply by 100 to find the percentage.


P(x<4)\text{ \%}=0.26599*100
P(x<4)\text{ \%}=26.599\text{ \%}
P(x<4)\text{ \%}=27\text{ \%}

Final answer:

27 % of the time will the police take 4 minutes or less to arrive.

User Hamsteyr
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