Question

The length of time it takes college students 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 take between 2.0 and 4.5 minutes to find a parking spot in the library lot. Report the probability to 3 decimal places.

Answer 1

Answer:

0.774 = 77.40% probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.5, \sigma = 1$

Find the probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot.

pvalue of Z when X = 4.5 subtracted by the pvalue of Z when X = 2. So

X = 4.5

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

$Z = \frac{4.5 - 3.5}{1}$

$Z = 1$

$Z = 1$ has a pvalue of 0.841

X = 2

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

$Z = \frac{2 - 3.5}{1}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.067

0.841 - 0.067 = 0.774

0.774 = 77.40% probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot.

Answer 2

Answer:

The probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot is 0.775.

Step-by-step explanation:

We are given that the length of time it takes college students 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 college students to find a parking spot in the library parking

The z-score probability distribution is given by ;

                Z = $\frac{ X-\mu}{\sigma}} }$ ~ N(0,1)

where, $\mu$ = mean time = 3.5 minutes

            $\sigma$ = standard deviation = 1 minute

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot is given by = P(2.0 minutes < X < 4.5 minutes)

P(2.0 minutes < X < 4.5 minutes) = P(X < 4.5 minutes) - P(X $\leq$ 2 minutes)

   P(X < 4.5) = P( $\frac{ X-\mu}{\sigma}} }$ < $\frac{ 4.5-3.5}{1}} }$ ) = P(Z < 1) = 0.84134  {using z table}

   P(X $\leq$ 2) = P( $\frac{ X-\mu}{\sigma}} }$$\leq$  $\frac{ 2-3.5}{1}} }$ ) = P(Z $\leq$ -1.50) = 1 - P(Z < 1.50)

                                              = 1 - 0.93319 = 0.06681                                    

The above probability is calculated using z table by looking at value of x = 1.50 in the z table which have an area of 0.93319.

Therefore, P(2.0 minutes < X < 4.5 minutes) = 0.84134 - 0.06681 = 0.775

Hence, probability that a randomly selected college student will take between 2.0 and 4.5 minutes to find a parking spot in the library lot is 0.775.