Area Under Normal Curve Worksheet Answers 5nd9

Area Under Normal Curve Worksheet Answers Assume that 𝑥 is normally distributed with a specified mean and standard deviation. Find the indicated probabilities : 1. 𝑃(3 ≤ 𝑥 ≤ 6); 𝜇 = 4 ; 𝜎 = 2 𝑧(𝑥 = 3) =

3−4 2

= −0.5

𝑧(𝑥 = 6) =

6−4 2

= 1.0

𝑃(−0.5 ≤ 𝑧 ≤ 1.0) = 𝑃(𝑧 = 1.0) − 𝑃(𝑧 = −0.5) = 0.8413 − 0.3085 = 0.5328

2. 𝑃(10 ≤ 𝑥 ≤ 26); 𝜇 = 15 ; 𝜎 = 4 𝑧(𝑥 = 10) =

10−15 4

= −1.25

𝑧(𝑥 = 26) =

26−15 4

= 2.75

𝑃(−1.25 ≤ 𝑧 ≤ 2.75) = 𝑃(𝑧 = 2.75) − 𝑃(𝑧 = −1.25) = 0.9971 − 0.1056 = 0.8915

3. 𝑃(8 ≤ 𝑥 ≤ 12); 𝜇 = 15 ; 𝜎 = 3.2 𝑧(𝑥 = 8) =

8−15 3.2

= −2.19

𝑧(𝑥 = 12) =

12−15 3.2

= −0.94

𝑃(−2.19 ≤ 𝑧 ≤ −0.94) = 𝑃(−0.94) − 𝑃(−2.19) = 0.1736 − 0.0143 = 0.1593

4. 𝑃(40 ≤ 𝑥 ≤ 47); 𝜇 = 50 ; 𝜎 = 15 𝑧(𝑥 = 40) =

40−50 15

= −0.67

𝑧(𝑥 = 47) =

47−50 15

= −0.2

𝑃(−0.67 ≤ 𝑥 ≤ −0.2) = 𝑃(𝑧 = −0.2) − 𝑃(𝑧 = −0.67) = 0.4207 − 0.2514 = 0.1693

5. 𝑃(𝑥 ≥ 120); 𝜇 = 100 ; 𝜎 = 15 𝑧(𝑥 = 120) =

120 − 100 = 1.33 15

𝑃(𝑧 ≥ 1.33) = 1 − 𝑃(𝑧 = 1.33) = 1 − 0.9082 = 0.0918

Find the 𝑧 − 𝑣𝑎𝑙𝑢𝑒 described and sketch the area on the given curve. 6. Find 𝑧 such that 6% of the standard normal curve lies to the left of 𝑧. 6% = 0.0600 From the table z = -1.55 at 0.0606 z = -1.56 at 0.0594 Right in between so 𝑧 = −1.555

-1.555

7. Find 𝑧 such that 55% of the standard normal curve lies to the left of 𝑧. 55% = 0.5500 From the table z = 0.12 at 0.5478 Z = 0.13 at 0.5517 0.5500 is closer to 0.5517 so 𝑧 = 0.13 0.13

8. Find 𝑧 such that 8% of the standard normal curve lies to the right of 𝑧. 8% to the right = 1 – 0.0800 = 0.9200 From the table z = 1.40 at 0.9192 z = 1.41 at 09207 0.9200 is closer to 0.9207 so 𝑧 = 1.41 1.41

9. Find 𝑧 such that 95% of the standard normal curve lies to the right of 𝑧. 95% to the right = 1 – 0.9500 = 0.0500 From the table z = - 1.64 at 0.0505 z = - 1.65 at 0.0495 Right in between so 𝑧 = −1.645 -1.645

10.

Find 𝑧 such that 98% of the standard normal curve lies between −𝑧 𝑎𝑛𝑑 𝑧. 1−0.9800 2

=

0.02 2

= 0.0100

From the table z = - 2.32 at 0.0102 z = - 2.33 at 0.0099 0.0100 is closer to 0.0099 so 𝑧 = −2.33 98% of the data will fall between −2.33 ≤ 𝑧 ≤ 2.33 11.

2.33

-2.33

Find 𝑧 such that 60% of the standard normal curve lies between −𝑧 𝑎𝑛𝑑 𝑧. 1 − 0.6000 0.4000 = = 0.2000 2 2

From the table z = - 0.84 at 0.2005 z = - 0.85 at 0.1977 0.2000 is closer to 0.2005 so 𝑧 = −0.84 60% of the data will fall between −0.84 ≤ 𝑧 ≤ 0.84

-0.84

0.84

12. A person’s blood glucose level and diabetes are closely related. Let 𝑥 be a random variable measured in milligrams of glucose per deciliter of blood. After a 12 – hour fast, the random variable 𝑥 will have a distribution that is approximately normal with a mean 𝜇 = 85 and a standard deviation 𝜎 = 25. After 50 years, these tend to increase. Find the probability that, an adult under 50 years of age after a 12 – hour fast will have : a) 𝑥 > 60

𝑧=

𝑥−𝜇 𝜎

=

60−85 25

= −1.0

𝑃(𝑧 > 1.0) = 1 − 𝑃(𝑧 = −1.0) = 1 − 0.1587 = 0.8413 b) 𝑥 < 110

𝑧=

110−85 25

= 1.0

𝑃(𝑧 < 1.0) = 0.8413 ( from the table )

c) 60 < 𝑥 < 110 𝑃(−1.0 < 𝑧 < 1.0) = 𝑃(𝑧 = 1.0) − 𝑃(𝑧 = −1.0) = 0.84130 − 0.1587 = 0.6826 ** z – scores from above

13. Quick Start makes 12 – volt car batteries. After many years of product testing, the company knows the average life of a Quick Start battery is normally distributed, with a mean of 45 months and a standard deviation of 8 months. a) If Quick Start fully refunds any batteries that fail within a 36 – month period of purchase, what percentage of batteries will the company expect to replace ? 𝑧=

36 − 45 = −1.13 8

𝑃(𝑥 < 36 𝑚𝑜𝑛𝑡ℎ𝑠 ) = 𝑃(𝑧 < −1.13) = 0.1292 The company will expect to replace 13% of the batteries b) If Quick Start does not want to make refunds for more than 10% of its batteries under full – refund guarantee policy, for how long should the company guarantee the batteries ? ( round to nearest month ) 10% = 0.1000

which coverts to 𝑧 = −1.28

𝑥 = 𝑧𝜎 + 𝜇 = −1.28(8) + 45 = −10.24 + 45 = 34.76 𝑜𝑟 35 The company should guarantee their batteries for 35 months 14. Do the following data sets have a normal curve ? Use Person’s index. a) Mean = 30.2 , standard deviation = 1.8 , median = 30 𝑃𝑒𝑎𝑟𝑠𝑜𝑛′ 𝑠 𝑖𝑛𝑑𝑒𝑥 =

3(𝑚𝑒𝑎𝑛 − 𝑚𝑒𝑑𝑖𝑎𝑛) 3(30.2 − 30) 3(0.2) = = = 0.333 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣 1.8 1.8

Data has a normal curve --- index value falls between -1 and 1 b) Mean = 134.3 , standard deviation = 40.6 , median = 150 𝑃𝑒𝑎𝑟𝑠𝑜𝑛′ 𝑠 𝑖𝑛𝑑𝑒𝑥 =

3(𝑚𝑒𝑎𝑛 − 𝑚𝑒𝑑𝑖𝑎𝑛) 3(134.3 − 150) 3(−15.7) = = = −1.2 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣 40.6 40.6

Data does not have a normal curve --- index value falls outside -1 to 1 c) Mean = 86 , standard deviation = 6.2 , median = 84 3(𝑚𝑒𝑎𝑛 − 𝑚𝑒𝑑𝑖𝑎𝑛) 3(86 − 84) 3(2) = = = 0.97 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣 6.2 6.2 Data has a normal curve --- index value falls between -1 and 1

𝑃𝑒𝑎𝑟𝑠𝑜𝑛′ 𝑠 𝑖𝑛𝑑𝑒𝑥 =