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1.
Use technology or a z-score table to answer the question.

The expression P(z<1.45) represents the area under the standard normal curve below the given value of z.

What is P(z<1.45)?


0.0735

0.0749

0.9251

0.9265


2.Use technology or a z-score table to answer the question.

The scores on a standardized test are normally distributed with a mean of 500 and a standard deviation of 60. Jake scored 520 on the test.

What percent of students scored below Jake?

Round your answer to the nearest whole number.


33%

57%

63%

72%


3.Use technology or a z-score table to answer the question.

Lengths of newborn girls are normally distributed with a mean of 49.2 cm and a standard deviation of 1.8 cm. Consider a group of 2000 newborn girls.

Approximately how many girls will be 51 cm or shorter?


1683

1856

1928

1964


4.The number 0.9967 represents the area under the standard normal curve below a particular z-score.

What is the z-score?



Enter your answer, as a decimal to the nearest hundredth, in the box.

5.Use technology or a z-distribution table to find the indicated area.

Suppose ages of cars driven by company employees are normally distributed with a mean of 8 years and a standard deviation of 3.2 years.

Approximately 75% of cars driven by company employees are older than what age?


2.1

4.8

5.9

10.2

Respuesta :

haileymcmullian

Answer:

1.) 0.9265

2.) 63%

3.) 1683

4.) 5.9

I hope this helped and give me brainiest please

Answer:

Step-by-step explanation:

1) From standard distribution table we find that

P(Z<1.45) = 0.5+0.4265

=0.9265 Option d

2) Given that X, The scores on a standardized test is N(500,60)

percent of students who scored below Jake

[tex]=100*P(X<520)\\=100*P(Z<\frac{20}{60} )\\=100(0.5+0.1293)\\=62.93%\\=63%[/tex]

3) X, lengths of new born girls is N(49.2, 1.8)

n = 2000

P(X<51) = P(Z<1) =0.84

No of girls =2000(0.84) =1683

4) P(Z<2.72) is the answer

5)X, ages of cars driven is N(8,3.2)

75% correspond to z=0.675

X = 8+0.675(3.2) =10.16=10.2

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