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aivan3 [116]
3 years ago
12

At a large department store, the average number of years of employment for a cashier is 5.7 with a standard deviation of 1.8 yea

rs, and the distribution is approximately normal. If an employee is picked at random, what is the probability that the employee has worked at the store for over 10 years? 99.2% 0.8% 49.2% 1.7%
Mathematics
1 answer:
Mandarinka [93]3 years ago
5 0

Answer:

option 0.8%

Step-by-step explanation:

Data provided in the question:

Mean = 5.7 years

Standard deviation, s = 1.8 years

Now,

P(the employee has worked at the store for over 10 years)

= P(X > 10 years)

= P (Z > \frac{X-Mean}{\sigma})

or

= P (Z > \frac{10-5.7}{1.8})

= P (Z > 2.389 )

or

= 0.008447     [from standard  z table]

or

= 0.008447 × 100% = 0.84% ≈ 0.8%

Hence,

the correct answer is option 0.8%

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Refer to the attached image.

Given: m \angle A = 65^\circ and measure of exterior angle at C = 117^\circ.

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By applying exterior angle property of the triangle which states:

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So, exterior angle C = m \angle A + m \angle B

117^\circ = 65^\circ+ m \angle B

m \angle B = 52^\circ

Now, applying angle sum property in triangle ABC which states:

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m \angle A + m \angle B + m \angle BCA = 180^\circ

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Therefore, the measure of m \angle B = 52^\circ and m \angle BCA =63^\circ.

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