Question

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.73 inch. The lower and upper specification limits under which the ball bearing can operate are 0.715 inch​ (lower) and 0.745 inch​ (upper). Past experience has indicated that the actual diameter of the ball bearings is approximately normally​ distributed, with a mean of 0.734 inch and a standard deviation of 0.006 inch. Suppose a random sample of 22 ball bearings are selected. Complete parts​ (a) through​ (e). What is the probability that the sample mean is between the target and the population mean of 0.734​? What is the probability that the sample mean is between the lower specification limit and the​ target? What is the probability that the sample mean is greater than the upper specification​ limit? What is the probability that the sample mean is less than the lower specification​ limit? The probability is 92​% that the sample mean diameter will be greater than what​ value?

Answer 1

1)

μ=0.734, σ=.006, n=22

We need to compute Pr(0.730≤Xˉ≤0.734).

The corresponding z-values needed to be computed are:

Therefore, the following is obtained:

2)

μ=0.734, σ=.006, n=22

We need to compute Pr(0.715≤Xˉ≤0.730).

The corresponding z-values needed to be computed are:

Therefore, the following is obtained:

3)

μ=0.734, σ=.006, n=22

We need to compute Pr(Xˉ≥0.745).

The corresponding z-value needed to be computed is:

Therefore, we get that

4)

μ=0.734, σ=.006, n=22

We need to compute Pr(Xˉ≤0.715).

The corresponding z-value needed to be computed:

Therefore,

5)

Pr [ X > x ] = 0.92 which is same as Pr[X <= x ] = 0.08

μ=0.734, σ=0.006

We need to find a score x so that the corresponding cumulative normal probability is equal to 0.08.

Mathematically, x is such that:

Pr(X≤x)=0.08

The corresponding z score so that the cumulative standard normal probability distribution is 0.08 is

z_c = - 1.4051

Hence, the X score associated with the 0.08 cumulative probability is

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