Question

Marcella takes a shower every morning when she gets up. Her time in the shower varies according to a Normal distribution with mean $4.5$minutes and standard deviation $0.9$minutes.

a. Find the probability that Marcella’s shower lasts between $3$and $6$minutes on a randomly selected day.

b. If Marcella took a $7$minute shower, would it be classified as an outlier by the $1.5IQR$rule? Justify your answer.

c. Suppose we choose $10$days at random and record the length of Marcella’s shower each day. What’s the probability that her shower time is $7$minutes or greater on at least $2$of the days?

d. Find the probability that the mean length of her shower times on these 10 days exceeds$5$ minutes.

Short answer

(a) The probability is $0.9050.$

(b) The 7-minute shower would be classed as an exception because $2.78$is higher than $2.68$

(c) The probability is $0.000323$

(d) The probability that the mean length of her shower times is$0.0392$.

Step-by-step solution

Part (a) Step 1: Given information

The given normal distribution has

Mean$=4.5min$

Standard deviation$=0.9min$

Part (a) Step 2: Explanation

The formula to find probability is

$z=\frac{x-\mu }{\sigma }$

Let $X$be the length of the researcher's shower on a randomly chosen day, with a mean of $4.5$minutes and a standard and standard deviation of $0.9$minutes.

Finding the $z-$value for the $P(3<X<6)$

Identifying the normal probability and associating them

$z_l=\frac{3-4.5}{0.9}$

$$\begin{gathered} =-1.67 \\ {p}_l=0.0475 \end{gathered}$$

$$\begin{gathered} z_u=\frac{6-4.5}{0.9} \\ =1.67 \\ \Rightarrow p_u=0.9525 \end{gathered}$$

To find the probability, find the difference in probabilities:

$$\begin{gathered} p_u-p_l=0.9525-0.0475 \\ =0.9050 \end{gathered}$$

Part (b) Step 1: Given information

The given normal distribution has

Mean$=4.5min$

Standard deviation$=0.9min$

Part (b) Step 2: Explanation

The formula used is

$z=\frac{x-\mu }{\sigma }$

$25th$percentile $Q_1$

Finding the $z-$score that corresponds to a probability of $0.25$in the normal probability table, it is discovered that the nearest probability is $0.2154$, which is located in the row $-0.6$and column $0.07$of the normal probability table, and so the corresponding $z-$score is

$$\begin{gathered} =-0.6+0.07 \\ =-0.67 \end{gathered}$$

$75thpercentile{Q}_3$

Finding the $z-$score that corresponds to a probability of $0.75$in the normal probability table, it is discovered that the nearest probability is $0.7486$, which is found in the row $0.6$and column $0.07$of the normal probability table, and thus the corresponding $z-$score is

$$\begin{gathered} =0.6+0.07 \\ =0.67 \end{gathered}$$

The inter quartile range $IQR$is

$$\begin{gathered} IQR=Q_3-Q_1 \\ =0.67-(-0.67) \\ =1.34 \end{gathered}$$

Outliers are

$$\begin{gathered} Q_3+1.5IQR=0.67+1.5(1.34) \\ =2.68 \\ {Q}_1-1.5IQR=-0.67-1.5\left(134\right) \\ =-2.68 \end{gathered}$$

All z-scores that are below $-2.68$or over $2.68$are considered outliers.

The z-score is a measure of how well something works.

$$\begin{gathered} z=\frac{7-4.5}{0.9} \\ =2.78 \end{gathered}$$

The 7-minute shower would be classed as an exception because $2.78$ is higher than$2.68$

Part (c) Step 1: Given information

The given normal distribution has

Mean$=4.5min$

Standard deviation$=0.9min$

Part (c) Step 2: Explanation

The formula is $z=\frac{x-\mu }{\sigma }$

The multiplication and complement rules are

$$\begin{gathered} P(AandB)=P\left(A\right)\times P\left(B\right) \\ P(notA)=1-P\left(A\right) \end{gathered}$$

From part (b) the $z-$score is $2.78$

$$\begin{gathered} P(X>7)=P(Z>2.78) \\ =P(Z<-2.78) \\ =0.0027 \end{gathered}$$

$$\begin{gathered} P(X\le 7)=P(Z<2.78) \\ =0.9973 \end{gathered}$$

Applying the rule

$P(Atleast2of10days>7)=1-P(0of10days>7)-P(1of10days>7)$

$$\begin{gathered} =1-0.{9973}^{10}-10\times 0.{9973}^9\times 0.0027 \\ =0.000323 \end{gathered}$$

Part (d) Step 1: Given information

The given normal distribution has

Mean$=4.5min$

Standard deviation$=0.9min$

Part (d) Step 2: Explanation

Here the formula used is

$z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$

The $z-$score is

$$\begin{gathered} =\frac{5-4.5}{0.9/\sqrt{10}} \\ =1.76 \end{gathered}$$

Then the associating probability is

$$\begin{gathered} P(X>5)=P(Z>1.76) \\ =P(Z<-1.76) \\ =0.0392 \end{gathered}$$