A battery manufacturer claims that the lifetime of a certain type of battery has a population mean of 40 hours and a standard deviation of 5 hours. Let X represent the mean lifetime of the batteries in a simple random sample of size 100.

If the claim is true, what is P(X(bar) < 36.7)?

Whenever I work out the problem I keep getting a P that's much too large to be correct. A step in the right direction would be very much appreciated!

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hafsaabdulhai hafsaabdulhai · Answer 1 · 2020-03-07T09:13:30+01:00

Answer:

0.25463

Step-by-step explanation:

z=(X-mean)/standard deviation

Here X=36.7,

z= (36.7-40)/5

z = -0.66

From look-up table of normal distribution (see attachment) , at z= -0.66

P(X(bar)<36.7) = 0.25463

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hamzaahmeds hamzaahmeds · Answer 2 · 2020-03-07T11:05:35+01:00

Answer:

The probability is found to be 0.2546.

Step-by-step explanation:

This is the case of normal distribution. The data given is:

Mean = μ = 40 hours

Standard Deviation = σ = 5 hours

X = 36.7 hours

Therefore, from the formula of normal distribution, we calculate Z, firt. The formula is as follows:

Z = (X - μ)/σ

Z = (36.7 - 40)/5

Z = -0.66

Therefore,

P (X < 36.7) = P (Z < -0.66)

Hence, from the table attached, we find the probability of z < -0.66, through normal distribution graph.

P (X < 36.7) = 0.2546