\(\displaystyle Z_{2.4}=.9918\)

\(\displaystyle Z_{1.5}=.9332\)

\(\displaystyle .9918-.9332=.0586\)

Alex - 

\(\displaystyle Z=\left ( 35-27\right )/5 = 1.6 = .945\) on the z-table

Noah - 

\(\displaystyle Z=\left ( 82-70\right )/8=1.5=.933\) on the z-table

\(\displaystyle Z_{12}=\left ( 12-19\right )\div6=-1.17\)

\(\displaystyle p_{-1.17}=.121\)

\(\displaystyle Z_{25}=\left ( 25-19\right )\div6=1\)

\(\displaystyle p_{1}=.8413\)

\(\displaystyle P\left ( 12< x< 25\right )=.8413 - .121 = .7203\)  

\(\displaystyle P(X=3)=\frac{(\lambda t)^n}{n!}e^{-\lambda t}=\frac{(4\cdot 1)^3}{3!}e^{-4\cdot 1}=19.5\%\)

The Z score for a normal distribution at the \(\displaystyle 25th\) percentile is \(\displaystyle z = -0.675.\) So  \(\displaystyle P(z < -0.675) \approx 0.25\), which can be found on the normal distribution table. The mass of tomatoes in the \(\displaystyle 25th\) percentile of all tomatoes  is \(\displaystyle 0.675\) standard deviations below the mean, so the mass is \(\displaystyle 45g - 0.675 \ast 4g \approx 42.3g\).

First, we use our normal distribution table to find a p-value for a z-score greater than 0.50.

Our table tells us the probability is approximately,

\(\displaystyle P(z>0.50)=0.6915\).

Next we use our normal distribution table to find a p-value for a z-score greater than 1.23.

Our table tells us the probability is approximately,

\(\displaystyle P(z< 1.23)=0.8907\).

We then subtract the probability of z being greater than 0.50 from the probability of z being less than 1.23 to give us our answer of,

\(\displaystyle 0.8907-0.6915=0.1992\).

First, we use the table to look up a p-value for z > -1.22.

This gives us a p-value of,

\(\displaystyle P(z>-1.22)=0.1112\).

Next, we use the table to look up a p-value for z > 1.59.

This gives us a p-value of,

\(\displaystyle P(z< 1.59)=0.9441\).

Finally we subtract the probability of z being greater than -1.22 from the probability of z being less than 1.59 to arrive at our answer of,

\(\displaystyle 0.9441-0.1112=0.8329\).

When we get this type of problem, first we need to calculate a z-score that we can use in our table.

To do that, we use our z-score formula:

 \(\displaystyle Z = \frac{(X - \mu)}{\sigma}\)

where,

\(\displaystyle \\X=score=940 \\ \mu=mean=850 \\ \sigma = sample\ standard\ deviation=100\)

Plugging into the equation we get:

\(\displaystyle Z=\frac{(940 - 850)}{100}=\frac{90}{100}=0.90\)

We then use our table to look up a p-value for z > 0.9. Since we want to calculate the probability of students who earned a higher score than Gabbie we need to subtract the P(z<0.9) to get our answer.

\(\displaystyle \\P(z>0.90)=1-P(z< 0.90) \\P(z>0.90) =1-0.8159 \\P(z>0.90)=0.1841\)

The two main parameters of the normal distribution are \(\displaystyle \mu\) and \(\displaystyle \sigma\). \(\displaystyle \mu\) is a location parameter which determines the location of the peak of the normal distribution on the real number line. \(\displaystyle \sigma\) is a scale parameter which determines the concentration of the density around the mean. Larger \(\displaystyle \sigma\)'s lead the normal to spread out more than smaller \(\displaystyle \sigma\)'s. 

The mean determines where the normal distribution lies on the real number line, while the variance determines the spread of the distribution.