Question

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch. What is the probability that a ball bearing is:___________.
a. between the target and the actual mean?
b. between the lower specification limit and the target?
c. above the upper specification limit?d. below the lower specification limit?

Answer 1

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal($\mu=0.753,\sigma^{2} =0.004^{2}$)

The z-score probability distribution for normal distribution is given by;

                                Z  =  $\frac{X-\mu}{\sigma} } }$  ~ N(0,1)

where, $\mu$ = population mean = 0.753 inch

           $\sigma$ = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X $\leq$ 0.75 inch)

      P(X < 0.753) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.753-0.753}{0.004} } }$ ) = P(Z < 0) = 0.50

      P(X $\leq$ 0.75) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.75-0.753}{0.004} } }$ ) = P(Z $\leq$ -0.75) = 1 - P(Z < 0.75)

                                                             = 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the  lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X $\leq$ 0.74 inch)

      P(X < 0.75) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.75-0.753}{0.004} } }$ ) = P(Z < -0.75) = 1 - P(Z $\leq$ 0.75)

                                                            = 1 - 0.7734 = 0.2266

      P(X $\leq$ 0.74) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.74-0.753}{0.004} } }$ ) = P(Z $\leq$ -3.25) = 1 - P(Z < 3.25)

                                                             = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

      P(X > 0.76) = P( $\frac{X-\mu}{\sigma} } }$ > $\frac{0.76-0.753}{0.004} } }$ ) = P(Z > -1.75) = 1 - P(Z $\leq$ 1.75)

                                                            = 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

      P(X < 0.74) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.74-0.753}{0.004} } }$ ) = P(Z < -3.25) = 1 - P(Z $\leq$ 3.25)

                                                            = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

Answer 2

Using the normal distribution, it is found that there is a

a) 0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.

b) 0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.

c) 0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.

d) 0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 0.753 inch, hence $\mu = 0.753$.
• Standard deviation of 0.004 inch, hence $\sigma = 0.004$

Item a:

This probability is the p-value of Z when X = 0.753 subtracted by the p-value of Z when X = 0.75.

X = 0.753:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.753 - 0.753}{0.004}$

$Z = 0$

$Z = 0$ has a p-value of 0.5

X = 0.75:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.75 - 0.753}{0.004}$

$Z = -0.75$

$Z = -0.75$ has a p-value of 0.2266

0.5 - 0.2266 = 0.2734

0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.

Item b:

This probability is the p-value of Z when X = 0.75 subtracted by the p-value of Z when X = 0.74, hence:

X = 0.75:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.75 - 0.753}{0.004}$

$Z = -0.75$

$Z = -0.75$ has a p-value of 0.2266

X = 0.74:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.74 - 0.753}{0.004}$

$Z = -3.25$

$Z = -3.25$ has a p-value of 0.0006.

0.2266 - 0.0006 = 0.226

0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 0.76, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.76 - 0.753}{0.004}$

$Z = 1.75$

$Z = 1.75$ has a p-value of 0.9599.

1 - 0.9599 = 0.0401

0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.

Item d:

This probability is the p-value of Z when X = 0.74, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.74 - 0.753}{0.004}$

$Z = -3.25$

$Z = -3.25$ has a p-value of 0.0006.

0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.

A similar problem is given at https://brainly.com/question/24663213