Question

Suppose that the scores of bowlers in particular league follow a normal distribution such that the standard deviation of the population is 6. Find the 95% confidence interval of the mean score for all bowlers in this league, using the accompanying data set of 10 random scores. Round your answers to two decimal places and use ascending order. Score 86 86 93 88 98 107 93 75 89

Answer 1

Answer:

A 95% confidence interval for the population mean score for all bowlers in this league is [86.64, 94.48].

Step-by-step explanation:

Since in the question only 9 random scores are given, so I am performing the calculation using 9 random scores.

We are given that the scores of bowlers in particular league follow a normal distribution such that the standard deviation of the population is 6.

The accompanying data set of 9 random scores in ascending order is given as; 75, 86, 86, 88, 89, 93, 93, 98, 107

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                             P.Q.  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean score = $\frac{\sum X}{n}$ = $\frac{815}{9}$ = 90.56

            $\sigma$ = population standard deviation = 6

            n = sample of random scores = 9

            $\mu$ = population mean score for all bowlers

Here for constructing a 95% confidence interval we have used a One-sample z-test statistics because we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =  [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

                                          = [ $90.56-1.96 \times {\frac{6}{\sqrt{9} } }$ , $90.56+1.96 \times {\frac{6}{\sqrt{9} } }$ ]

                                          = [86.64 , 94.48]

Therefore, a 95% confidence interval for the population mean score for all bowlers in this league is [86.64, 94.48].