The heights, in inches, for 120 people (male and female) have been measured. The results are presented:

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66.88	67.88	69.6	60.16	64.24	65.12	64.23	59.66	67.52	69.85	69.59	65.59
67.79	73.41	67.65	68.22	64.91	69.03	66.47	65.57	74.26	65.57	67.22	65.95
64.82	68.27	65.96	66.49	67.19	67.24	65.96	70.25	68.33	63.44	70.76	69.87
65.26	65.56	64.66	69.74	66.45	63.69	64.98	66.04	65.78	65.27	65.15	70.04
62.2	68.37	70.32	66.74	62.26	68.62	68.39	67.48	66.06	67.66	68.06	69.05
68.95	61.95	68.5	64.99	66.57	64.94	65.3	69.68	66.95	68.43	65.49	62.83
63.96	63.7	61.42	61.58	65.81	65.8	69.53	69.25	64.14	68.59	66.49	64.6
67.52	63.96	67.04	72.39	74.5	66.2	60.73	67.42	69.74	68.3	65.12	64.39
66.51	66.9	65.82	66.29	64.77	64.58	70.02	69.99	68.72	64.76	63.64	63.29
67.18	71.03	66.16	66.11	65.85	67.8	64.72	63.95	65.6	61.32	66.55	63.57

a)Calculate the mean and standard deviation for the sample. Give your answers to 2 decimal places.

sample mean =

sample standard deviation =

b)Find the proportion of heights that are within 1 standard deviation of the sample mean and also the proportion that are within 2 standard deviations of the sample mean. Use theunroundedvalues for the mean and standard deviation when doing this calculation. Give your answers as decimals to 2 decimal places.

Proportion of heights within 1 standard deviation of the mean =

Proportion of heights within 2 standard deviations of the mean =

c)Calculate the standardized value for the value 66.

z-score for the value 66 =

d)Calculate the probability that a standard normal random variable Z takes a values less than the standardized value calculated in part c). You may find thisstandard normal tableuseful. Give your answer as a decimal to 4 decimal places.

Probability Z less than standardized value =

e)Find the proportion of values in the sample that are less than or equal to 66. Give you answer as a decimal to 2 decimal places.

Proportion of values less than or equal to 66 =

f)Select the appropriate description for the data:

the data are APPROXIMATELY normal the data are CLEARLY not normal