If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. What type of problem is this? Poisson Random variable Sample mean Sample proportion Normal

Respuesta :

Answer:

a) The probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes = 0.309

b) Check the beginning of Explanation for which type of dustribution this is.

Step-by-step explanation:

A distribution with the mean given and the spread around the mean (standard deviation) given, is a normal distribution. Especially as it shows that the distribution is symmetric around the mean.

To find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes.

We first standardize '3 minutes'.

The standardized score is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (3 - 3.5)/1 = - 0.5

probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes.

P(x < 3) = P(z < -0.5)

We'll use data from the normal probability table for these probabilities

P(x < 3) = P(z < -0.5) = 1 - P(z ≥ - 0.5) = 1 - P(z ≤ 0.5) = 1 - 0.691 = 0.309.

Answer:

(a) P(X < 3) = 0.30854

(b) This problem is of Normal distribution.

Step-by-step explanation:

We are given that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute

Let X = length of time it takes a college student to find a parking spot in the library parking lot

So, X ~ N([tex]\mu=3.5,\sigma^{2} = 1^{2}[/tex])

The z score probability distribution is given by;

                Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)  

where, [tex]\mu[/tex] = population mean

           [tex]\sigma[/tex] = population standard deviation

(a) Probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes is given by = P(X < 3)

    P(X < 3) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{3-3.5}{1}[/tex] ) = P(Z < -0.5) = 1 - P(Z [tex]\leq[/tex] 0.5)

                                              = 1 - 0.69146 = 0.30854

(b) This problem is of Normal distribution a stated in the question above.