Question

mat 2170 worksheet 4.5 4.6 4.8 continuous distributions show all your work using formulas & sketching/shading area under distribution curve! 1) the average time a subscriber spends reading the wall street journal is 49 minutes. assume the standard deviation is 16 minutes and the distribution is normal. a) what is the probability a subscriber will spend at most 1 hour reading the journal? i. convert 1 hour to a standard normal z - score. ii. sketch the normal curve and shade the area underneath that represents the probability a subscriber will spend at most 1 hour reading the journal. iii. use a normal probability table and symmetry of the normal distribution curve to find correct probability. b) what is the probability a subscriber will spend more than 30 minutes reading the journal? (show work following the same steps as in part a.)

Explanation:

Step1: Calculate Z - score for part a

The formula for the Z - score is $Z=\frac{X-\mu}{\sigma}$, where $X$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation. Given $\mu = 49$ minutes, $\sigma=16$ minutes and $X = 60$ minutes (since 1 hour = 60 minutes). Then $Z=\frac{60 - 49}{16}=\frac{11}{16}=0.6875$.

Step2: Sketch the normal curve for part a

Draw a bell - shaped normal curve. Mark the mean $\mu = 49$ on the horizontal axis. Shade the area to the left of $X = 60$ (or $Z=0.6875$) under the curve. This shaded area represents the probability that a subscriber will spend at most 1 hour reading the journal.

Step3: Find the probability for part a

Using a standard normal probability table (Z - table), we look up the value corresponding to $Z = 0.6875$. The value in the table for $Z=0.69$ (rounding 0.6875) is approximately $0.7549$. So $P(X\leq60)=P(Z\leq0.6875)\approx0.7549$.

Step4: Calculate Z - score for part b

Using the Z - score formula $Z=\frac{X-\mu}{\sigma}$, with $\mu = 49$ minutes, $\sigma = 16$ minutes and $X = 30$ minutes. Then $Z=\frac{30 - 49}{16}=\frac{- 19}{16}=-1.1875$.

Step5: Sketch the normal curve for part b

Draw a bell - shaped normal curve. Mark the mean $\mu = 49$ on the horizontal axis. Shade the area to the right of $X = 30$ (or $Z=-1.1875$) under the curve.

Step6: Find the probability for part b

We know that $P(X > 30)=1 - P(X\leq30)$. First, we look up the value for $Z=-1.19$ (rounding - 1.1875) in the Z - table. The value for $Z=-1.19$ is approximately $0.1170$. So $P(X > 30)=1 - 0.1170 = 0.8830$.

Answer:

a. $P(X\leq60)\approx0.7549$
b. $P(X > 30)\approx0.8830$