The mean life of a television set is 119 months with a standard deviation of 13 months. If a sample of 67 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by more than 2 months? Round your answer to

The probability that the sample mean will differ from the true mean by more than two months is 0.10383.

Step-by-step explanation:

Hello!

The variable of interest is X: the lifespan of a television set.

The mean lifespan is μ= 119 months and the standard deviation is δ= 13 months.

A sample of n=65 televisions is taken.

There is no information about the distribution of the variable of interest, but, since the sample is greater than 30, you can apply the Central Limit Theorem and approximate the sampling distribution to normal:

X[bar]≈N(μ;δ²/n)

Thanks to this you can use an approximation of the standard normal distribution to calculate the asked probability:

Z= (X[bar]-μ)/(δ/√n) ≈ N(0;1)

You need to calculate the probability of the sample mean differ from the true mean by more than two months, symbolically: X[bar] ≥ μ + 2 months

P(X[bar]≥121)= 1 - P(X[bar]<121)

Now you have to standardize the value of X to obtain the corresponding Z value.

1 - P(Z<(121-119)/(13/√67))= 1 - P(Z<1.26) = 1 - 0.89617= 0.10383.

I hope this helps!

mark brainliest

step-by-step explanation:

2 tens + 6 ones

1 ten + 16 ones

0 tens + 26 ones