Question

In: Statistics and Probability

Shaver manufacturing, Inc. offers dental insurance to its employees. A recent study by the Human Resource...

Shaver manufacturing, Inc. offers dental insurance to its employees. A recent study by the Human Resource Director shows the annual cost per employee per year followed the normal distribution, with a mean of $1 280 and the standard deviation of $420 per year. In all questions start with sketching a normal curve. Cross-hatch requested area/probability. If you calculate z-value, round it to two decimal places.

What was the cost for the 15% of employees that incurred the highest dental expense?
What was the range of annual dental costs (in dollars) for middle 80% of all employees?
What is the probability the employee’s annual dental costs are not exceeding $1200?
If random sample of 50 employees is chosen, what is the probability the sample is having a mean of $1100 or more?
If random sample of 95 employees is chosen, what is the probability the sample is having a mean between $1300 and $1400?

Solutions

Expert Solution

Solution:-

Given that,

mean = = 1280

standard deviation = = 420

The z - distribution of the 15%

P(Z > z) = 15%

= 1 - P(Z < z ) = 0.15

= P(Z < ) = 1 - 0.15

P(Z < z ) = 0.85

P(Z < 1.04) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 420 + 1280

x = 559104

( b )

Middle 90%

= 1 - 80%

= 1 - 0.80 = 0.2

/2 = 0.1

1 - /2 = 1 - 0.1 = 0.9

Z/2 = Z0.1 = -1.28

Z1- /2 = Z 0.9 = 2.28

Using z-score formula,

x = z * +

x = -1.28 * 420 + 1280

x = 742.4

Using z-score formula,

x = z * +

x = 1.28 * 420 + 1280

x = 1817.6

Between 742 and 1818

( c )

P(x < 1200 )

= P[(x - ) / < ( 1200 - 1280 ) / 420 ]

= P(z < -0.19 )

Using z table,

= 0.4247

Probability = 0.4247

( d )

n = 50

= 1280

= / n = 420 / 50 = 59.3970

P( > 1100 ) = 1 - P( < 1100)

= 1 - P[( - ) / < ( 1100 - 1280 ) / 59.3970 ]

= 1 - P(z < -3.03)

Using z table,

= 1 - 0.0012

= 0.9988

Probability = 0.9988

( e )

n = 95

= 1280

= / n = 420 / 95 = 43.0911

P( 1300 < < 1400)

= P[( 1300 - 1280 ) / 43.0911 < ( - ) / < ( 1400 - 1280 ) / 43.0911 )]

= P( 0.46 < Z < 2.78 )

= P(Z < 2.78 ) - P(Z < 0.46 )

Using z table,

= 0.9973 - 0.6772

= 0.3201

Probability = 0.3201

orchestra answered 1 year ago

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