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Diano4ka-milaya [45]
3 years ago
12

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a stand

ard deviation of 2 minutes. Seventy percent of the time, it takes more than how many minutes to find a parking space? (Round your answer to two decimal places.)
Mathematics
1 answer:
erastovalidia [21]3 years ago
3 0

Answer:

z=0.524

And if we solve for a we got

a=5 +0.524*2=6.048 \approx 6.05 min

Step-by-step explanation:

Let X the random variable that represent the lenght time it takes to find a parking space at 9AM of a population, and for this case we know the distribution for X is given by:

X \sim N(5,2)  

Where \mu=5 and \sigma=2

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.3   (a)

P(X   (b)

As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.524

And if we solve for a we got

a=5 +0.524*2=6.048 \approx 6.05 min

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