The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs i

Answer:

Answer and Explanation:

We have:

Population mean,

μ

=

3

,

000

hours

Population standard deviation,

σ

=

696

hours

Sample size,

n

=

36

1) The standard deviation for the sampling distribution:

σ

¯

x

=

σ

√

n

=

696

√

36

=

116

2) By the central limit theorem, the sampling distribution's expected value matches the population mean.

Thus:

The expected value of the sampling distribution equals the population mean,

μ

¯

x

=

μ

=

3

,

000

The standard deviation of the sampling distribution,

σ

¯

x

=

116

The sampling distribution of

¯

x

is roughly normal due to a sample size greater than

30

.

3) The likelihood that the average lifespan of the sample falls between

2670.56

and

2809.76

hours:

P

(

2670.56

<

x

<

2809.76

)

=

P

(

2670.56

−

3000

116

<

z

<

2809.76

−

3000

116

)

=

P

(

−

2.84

<

z

<

−

1.64

)

=

P

(

z

<

−

1.64

)

−

P

(

z

<

−

2.84

)

=

0.0482

In Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)

4) The probability of the average life in the sample exceeding

3219.24

hours:

P

(

x

>

3219.24

)

=

P

(

z

>

3219.24

−

3000

116

)

=

P

(

z

>

1.89

)

=

0.0294

In Excel: =NORMSDIST(-1.89)

5) The likelihood that the sample's average life is lower than

3180.96

hours:

P

(

x

<

3180.96

)

=

P

(

z

<

3180.96

−

3000

116

)

=

P

(

z

<

1.56

)

=

0.9406