How to use tables of normal distribution - AP Statistics

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Question

Alex took a test in physics and scored a 35. The class average was 27 and the standard deviation was 5.

Noah took a chemistry test and scored an 82. The class average was 70 and the standard deviation was 8.

Show that Alex had the better performance by calculating -

Alex's standard normal percentile and

Noah's standard normal percentile

Answer

Alex -

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

Noah -

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

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Question

Find the area under the standard normal curve between Z=1.5 and Z=2.4.

Answer

$Z_{2.4}=.9918$

$Z_{1.5}=.9332$

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Question

When

$\mu=19$

and

$\sigma=6$

Find

$P\left ( 12< x< 25\right )$ .

Answer

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

$p_{-1.17}=.121$

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

$p_{1}=.8413$

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

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Question

Arrivals to a bed and breakfast follow a Poisson process. The expected number of arrivals each week is 4. What is the probability that there are exactly 3 arrivals over the course of one week?

Answer

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

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Question

The masses of tomatoes are normally distributed with a mean of grams and a standard deviation of grams. What mass of tomatoes would be the percentile of the masses of all the tomatoes?

Answer

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

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Question

Find

.

Answer

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,

.

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,

.

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,

.

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Question

Find

.

Answer

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

This gives us a p-value of,

.

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

This gives us a p-value of,

.

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,

.

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Question

Gabbie earned a score of 940 on a national achievement test. The mean test score was 850 with a sample standard deviation of 100. What proportion of students had a higher score than Gabbie? (Assume that test scores are normally distributed.)

Answer

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:

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

where,

$\X=score=940 \ \mu=mean=850 \ \sigma = sample\ standard\ deviation=100$

Plugging into the equation we get:

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

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.

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

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Question

Alex took a test in physics and scored a 35. The class average was 27 and the standard deviation was 5.

Noah took a chemistry test and scored an 82. The class average was 70 and the standard deviation was 8.

Show that Alex had the better performance by calculating -

Alex's standard normal percentile and

Noah's standard normal percentile

Answer

Alex -

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

Noah -

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

Compare your answer with the correct one above

Question

Find the area under the standard normal curve between Z=1.5 and Z=2.4.

Answer

$Z_{2.4}=.9918$

$Z_{1.5}=.9332$

Compare your answer with the correct one above

Question

When

$\mu=19$

and

$\sigma=6$

Find

$P\left ( 12< x< 25\right )$ .

Answer

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

$p_{-1.17}=.121$

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

$p_{1}=.8413$

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

Compare your answer with the correct one above

Question

Arrivals to a bed and breakfast follow a Poisson process. The expected number of arrivals each week is 4. What is the probability that there are exactly 3 arrivals over the course of one week?

Answer

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

Compare your answer with the correct one above

Question

The masses of tomatoes are normally distributed with a mean of grams and a standard deviation of grams. What mass of tomatoes would be the percentile of the masses of all the tomatoes?

Answer

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

Compare your answer with the correct one above

Question

Find

.

Answer

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,

.

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,

.

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,

.

Compare your answer with the correct one above

Question

Find

.

Answer

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

This gives us a p-value of,

.

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

This gives us a p-value of,

.

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,

.

Compare your answer with the correct one above

Question

Gabbie earned a score of 940 on a national achievement test. The mean test score was 850 with a sample standard deviation of 100. What proportion of students had a higher score than Gabbie? (Assume that test scores are normally distributed.)

Answer

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:

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

where,

$\X=score=940 \ \mu=mean=850 \ \sigma = sample\ standard\ deviation=100$

Plugging into the equation we get:

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

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.

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

Compare your answer with the correct one above

Question

Alex took a test in physics and scored a 35. The class average was 27 and the standard deviation was 5.

Noah took a chemistry test and scored an 82. The class average was 70 and the standard deviation was 8.

Show that Alex had the better performance by calculating -

Alex's standard normal percentile and

Noah's standard normal percentile

Answer

Alex -

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

Noah -

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

Compare your answer with the correct one above

Question

Find the area under the standard normal curve between Z=1.5 and Z=2.4.

Answer

$Z_{2.4}=.9918$

$Z_{1.5}=.9332$

Compare your answer with the correct one above

Question

When

$\mu=19$

and

$\sigma=6$

Find

$P\left ( 12< x< 25\right )$ .

Answer

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

$p_{-1.17}=.121$

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

$p_{1}=.8413$

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

Compare your answer with the correct one above

Question

Arrivals to a bed and breakfast follow a Poisson process. The expected number of arrivals each week is 4. What is the probability that there are exactly 3 arrivals over the course of one week?

Answer

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

Compare your answer with the correct one above

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