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!

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

the answer is:

the lateral area is equal to

to find the lateral area of the square pyramid, first we need to find the perimeter of the base, and then, find the lateral area using the following formulas:

where,

s, is the base length

slant height, is the height of a the side of the pyramid

we are given:

now, calculating the perimeter of the base, we have:

then, calculating the lateral area:

hence, we have that the lateral area is equal to

have a nice day!