關於Standard deviation同mean既問題..

👨🏻‍🏫 學術台數學

[匿名]

2009-01-07 5:21 am

A manufacturer of television sets has bought 7,500 LCD displays from an oversea supplier. The supplier claims that the life times of the displays are normally distributed with the mean life of 3,800 hours and the standard deviation is 360 hours.

(a) How many displays are expected to stop working in the first 3,500 hours? (5 marks)

(b) Determine the number of hours for which 12% of the displays ordered would be expected to stop working? (5 marks)

**{ PLEASE SHOW THE STEPS, THANKS ^^ }

更新1:

= 0.5 - 0.297633333 呢度點黎架? 唔明...

回答 (2)

Audrey Hepburn

2009-01-07 5:32 am

✔ 最佳答案

Let X be the random variable of the mean life of the displays.

X ~ N(3800 , 3602)

a. P(stop working in the first 3500 hours)

= P(X <= 3500)

= P[z <= (3500 - 3800)/360]

= P(z <= -5/6)

= 0.5 - 0.297633333

= 0.202366666

So, expected number of displays to stop working in the first 3500 hours

= 7500 X (0.202366666)

= 1517.75

b. Let c be the minimum time for which 12% of the displays to stop working.

P(X <= c) = 0.12

P[z <= (c - 3800)/360] = 0.12

(c - 3800)/360 = -1.175

c = 3377

So, the required time is 3377 hours.

參考: Myself~~~

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myisland8132

2009-01-07 5:36 am

(a)

μ=3800, σ=360

P(X<=3500)

=P(Z<=3500-3800)/360)

=P(Z<=-300/360)

=P(Z<=-0.833)

=0.2024

The expected number is 7500*0.2024=1518

(b)

Let the time is T

Then P(X

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